Solved Examples

How to Use This Section

You have finished the theory. Sections 1 to 10 took you from the cathode ray tube to the electronic configuration of copper, and each carried its own dozen worked examples. This section is different: it is one long problem set — 36 fully worked questions arranged from the simplest to the hardest — and it deliberately covers every in-text problem and every end-of-chapter exercise that boards, JEE Main and NEET keep recycling, plus four harder multi-step problems of the kind JEE likes to build from this chapter.

Structure of Atom is not conceptually hard; the arithmetic is what costs marks. Wavelengths left in nanometres inside $E = hc/\lambda$, a forgotten $Z^2$ in the Bohr energy, the electrons of the atom counted when the question asked about the ion, $l = 3$ written for $n = 3$ — those are the losses. Volume of practice, with units written at every line, is the only cure. So work these.

The working method

1. Cover the answer. Attempt the question first, on paper, with a pen. Reading a solution feels like learning and is not.
2. Compare your steps, not just your number. If you reached the right answer by a longer route, note the shorter one; if you reached the wrong answer, find the exact line where it went wrong.
3. Read the Watch out line. Most questions end with one. That is the transferable part; the numbers and the particular examples are disposable.

The constants used throughout

Key Point: $h = 6.626 \times 10^{-34}$ J s; $c = 3.0 \times 10^{8}\ \mathrm{m\ s^{-1}}$; $m_e = 9.1 \times 10^{-31}$ kg ($9.1094 \times 10^{-31}$ kg where the precision matters); $e = 1.602 \times 10^{-19}$ C; $N_A = 6.022 \times 10^{23}\ \mathrm{mol^{-1}}$; $R_H = 2.18 \times 10^{-18}$ J $= 13.6$ eV; $a_0 = 52.9$ pm; Rydberg constant $= 109{,}677\ \mathrm{cm^{-1}}$; $1\ \mathrm{eV} = 1.602 \times 10^{-19}$ J; and the exam shortcut $hc = 1240$ eV nm. A few answers are quoted to one more digit than the data justify, simply so that you can compare with your textbook's answer key.

The seven errors that cost the most marks in this chapter

Error	Where it bites	The fix
Leaving $\lambda$ in nm or Å inside $E = hc/\lambda$	Questions 11 to 16	Convert to metres first, every single time
Using the electron count instead of the proton count for an ion	Questions 2, 4	Protons $= Z$ always; electrons $= Z -$ charge
Forgetting $Z^2$ for $\mathrm{He^+}$ and $\mathrm{Li^{2+}}$	Questions 22, 23, 33	$E_n \propto Z^2/n^2$, $r_n \propto n^2/Z$
Writing $1/n_1^2 - 1/n_2^2$ with the levels the wrong way round	Questions 19 to 21	Put the smaller $n$ first; the wavenumber must come out positive
Mixing eV and joules in the photoelectric equation	Questions 15 to 18	Pick one unit before you subtract
Using $\Delta x \cdot \Delta p \geq h/2\pi$ instead of $h/4\pi$	Question 27	The exam form is $h/4\pi$
Letting $l$ reach $n$, or $m_l$ exceed $l$	Questions 28 to 30	$l$ runs $0$ to $n - 1$; $m_l$ runs $-l$ to $+l$

Solved Examples

Question 1: Protons, neutrons and electrons in a neutral atom

(a) Calculate the number of protons, neutrons and electrons in $^{80}_{35}\mathrm{Br}$. (b) How many neutrons and protons are there in the nuclei $^{13}_{6}\mathrm{C}$, $^{16}_{8}\mathrm{O}$, $^{24}_{12}\mathrm{Mg}$, $^{56}_{26}\mathrm{Fe}$ and $^{88}_{38}\mathrm{Sr}$?

Answer:

The symbol $^{A}_{Z}\mathrm{X}$ tells me everything. The bottom number $Z$ is the atomic number, which is the number of protons. The top number $A$ is the mass number, protons plus neutrons. Since it's a neutral atom, electrons equal protons.

(a) For $^{80}_{35}\mathrm{Br}$, $Z = 35$ and $A = 80$. $$\text{protons} = 35, \qquad \text{electrons} = 35, \qquad \text{neutrons} = A - Z = 80 - 35 = 45$$

(b) Same thing for each nucleus, neutrons $= A - Z$:

Nucleus	$Z$ (protons)	$A$	Neutrons $= A - Z$
$^{13}_{6}\mathrm{C}$	6	13	7
$^{16}_{8}\mathrm{O}$	8	16	8
$^{24}_{12}\mathrm{Mg}$	12	24	12
$^{56}_{26}\mathrm{Fe}$	26	56	30
$^{88}_{38}\mathrm{Sr}$	38	88	50

One thing I noticed: for the light ones $n \approx p$, but by iron and strontium the neutrons are clearly ahead. Heavier nuclei need extra neutrons to hold together against all the proton-proton repulsion.

Ans: (a) 35 protons, 45 neutrons, 35 electrons. (b) C: 6 p, 7 n; O: 8 p, 8 n; Mg: 12 p, 12 n; Fe: 26 p, 30 n; Sr: 38 p, 50 n.

Watch out: The only way to mess this up is mixing up which number is $A$ and which is $Z$. Bottom is protons, top is the total.

Question 2: From particle counts to the symbol, and electrons in molecular species

(a) The number of electrons, protons and neutrons in a species are 18, 16 and 16 respectively. Assign the proper symbol to the species. (b) Write the complete symbol for the atom with (i) $Z = 17$, $A = 35$; (ii) $Z = 92$, $A = 233$; (iii) $Z = 4$, $A = 9$. (c) Give the number of electrons in the species $\mathrm{H_2^+}$, $\mathrm{H_2}$ and $\mathrm{O_2^+}$.

Answer:

(a) The protons decide the element. 16 protons means $Z = 16$, so it's sulphur. Mass number is $A = p + n = 16 + 16 = 32$. There are 18 electrons but only 16 protons, so the species has 2 extra electrons and a charge of $-2$: $$^{32}_{16}\mathrm{S^{2-}}$$

(b) I just find the element from $Z$ and write the numbers on it. (i) $Z = 17$ is chlorine, $^{35}_{17}\mathrm{Cl}$; (ii) $Z = 92$ is uranium, $^{233}_{92}\mathrm{U}$; (iii) $Z = 4$ is beryllium, $^{9}_{4}\mathrm{Be}$.

(c) I add up the electrons of all the atoms first, then fix for the charge.

• $\mathrm{H_2}$: $2 \times 1 = 2$ electrons.
• $\mathrm{H_2^+}$: one electron has been taken away, so $2 - 1 = 1$ electron.
• $\mathrm{O_2^+}$: $\mathrm{O_2}$ has $2 \times 8 = 16$ electrons, minus one for the positive charge $= 15$ electrons.

Ans: (a) $^{32}_{16}\mathrm{S^{2-}}$. (b) $^{35}_{17}\mathrm{Cl}$, $^{233}_{92}\mathrm{U}$, $^{9}_{4}\mathrm{Be}$. (c) $\mathrm{H_2^+}$: 1, $\mathrm{H_2}$: 2, $\mathrm{O_2^+}$: 15.

Watch out: I write charge $=$ protons $-$ electrons, and the sign comes out right by itself: $16 - 18 = -2$.

Question 3: An atom with 29 electrons and 35 neutrons

An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.

Answer:

(i) It says "atom", so it's neutral and protons $=$ electrons $= 29$. $Z = 29$ is copper. The mass number is $A = 29 + 35 = 64$, so this is $^{64}_{29}\mathrm{Cu}$.

(ii) I fill 29 electrons in the aufbau order $1s\ 2s\ 2p\ 3s\ 3p\ 4s\ 3d$. $1s^2 2s^2 2p^6 3s^2 3p^6$ takes 18, so 11 are left for $4s$ and $3d$. If I fill blindly I get $4s^2 3d^9$.

But copper is one of the exceptions. A fully filled $3d^{10}$ is more stable than $3d^9 4s^2$ (more symmetry, more exchange energy), so one $4s$ electron shifts into $3d$: $$\mathrm{Cu}:\ 1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1 \quad \text{or} \quad [\mathrm{Ar}]\,3d^{10} 4s^1$$

Ans: (i) 29 protons ($^{64}_{29}\mathrm{Cu}$). (ii) $[\mathrm{Ar}]\,3d^{10} 4s^1$.

Watch out: Whenever a count lands on $Z = 24$ or $Z = 29$, it's a check on the two exceptions. Cr is $[\mathrm{Ar}]\,3d^5 4s^1$ and Cu is $[\mathrm{Ar}]\,3d^{10} 4s^1$; I nearly wrote $3d^9 4s^2$ for copper here.

Question 4: The "per cent more neutrons" puzzles

(a) An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol. (b) An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion. (c) An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.

Answer:

I set up the algebra once and reuse it. Let $p$ = protons, $n$ = neutrons, $e$ = electrons. Always $p + n = A$ and $e = p - (\text{charge})$. "$x\%$ more neutrons than electrons" just means $n = \left(1 + \dfrac{x}{100}\right)e$.

(a) Neutral atom, so $e = p$. That gives $n = 1.317p$, and $p + n = 81$: $$p + 1.317p = 81 \quad \Rightarrow \quad 2.317p = 81 \quad \Rightarrow \quad p = 34.96 \approx 35$$ So $n = 81 - 35 = 46$. $Z = 35$ is bromine: $^{81}_{35}\mathrm{Br}$.

(b) One negative charge means one extra electron, $e = p + 1$. The percentage is on electrons, so $n = 1.111e = 1.111(p + 1)$, and $p + n = 37$: $$p + 1.111(p + 1) = 37 \quad \Rightarrow \quad 2.111p = 35.889 \quad \Rightarrow \quad p = 17.0$$ So $e = 18$, $n = 37 - 17 = 20$. Quick check: $20/18 = 1.111$. $Z = 17$ is chlorine: $^{37}_{17}\mathrm{Cl^-}$.

(c) Three positive charges means three electrons missing, $e = p - 3$. So $n = 1.304(p - 3)$, and $p + n = 56$: $$p + 1.304p - 3.912 = 56 \quad \Rightarrow \quad 2.304p = 59.912 \quad \Rightarrow \quad p = 26.0$$ So $e = 23$, $n = 56 - 26 = 30$. Check: $30/23 = 1.304$. $Z = 26$ is iron: $^{56}_{26}\mathrm{Fe^{3+}}$.

Ans: (a) $^{81}_{35}\mathrm{Br}$ (b) $^{37}_{17}\mathrm{Cl^-}$ (c) $^{56}_{26}\mathrm{Fe^{3+}}$.

Watch out: In (b) and (c) the percentage is relative to the electrons of the ion, not the protons, so I have to write $e$ in terms of $p$ before substituting. $p$ should come out within a few hundredths of a whole number; if I get something like 17.4, I've used the wrong sign for the charge.

Question 5: Electrons that weigh one gram, and one mole of electrons

(i) Calculate the number of electrons which will together weigh one gram. (ii) Calculate the mass and charge of one mole of electrons.

Answer:

(i) One electron has mass $9.1094 \times 10^{-31}$ kg $= 9.1094 \times 10^{-28}$ g. I want to know how many of these add up to 1 g, so I divide: $$N = \frac{1\ \mathrm{g}}{9.1094 \times 10^{-28}\ \mathrm{g}} = 1.098 \times 10^{27}\ \text{electrons}$$

(ii) For the mass of a mole I multiply one electron's mass by Avogadro's number: $$m = 9.1094 \times 10^{-31}\ \mathrm{kg} \times 6.022 \times 10^{23}\ \mathrm{mol^{-1}} = 5.486 \times 10^{-7}\ \mathrm{kg\ mol^{-1}} = 0.549\ \mathrm{mg}$$

For the charge, each electron carries $-1.602 \times 10^{-19}$ C: $$q = -1.602 \times 10^{-19}\ \mathrm{C} \times 6.022 \times 10^{23} = -9.65 \times 10^{4}\ \mathrm{C}$$ That's one faraday (96,500 C), with a negative sign because it's electrons.

So a mole of electrons weighs about half a milligram, which is why we ignore electron mass in atomic masses, but it carries 96,500 C, which is why we can't ignore it in electrochemistry.

Ans: (i) $1.098 \times 10^{27}$ electrons. (ii) Mass $5.486 \times 10^{-7}$ kg (about 0.55 mg); charge $-9.65 \times 10^{4}$ C.

Question 6: Electrons, neutrons and protons in a given mass of substance

(i) Calculate the total number of electrons present in one mole of methane. (ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of $^{14}\mathrm{C}$. (Mass of a neutron $= 1.675 \times 10^{-27}$ kg.) (iii) Find (a) the total number and (b) the total mass of protons in 34 mg of $\mathrm{NH_3}$ at STP. Will the answer change if the temperature and pressure are changed?

Answer:

The route every time is mass, then moles, then molecules, then particles per molecule.

(i) One $\mathrm{CH_4}$ molecule has $6 + 4 \times 1 = 10$ electrons. One mole has $6.022 \times 10^{23}$ molecules: $$N_e = 10 \times 6.022 \times 10^{23} = 6.022 \times 10^{24}\ \text{electrons}$$

(ii) Molar mass of $^{14}\mathrm{C}$ is 14 g mol$^{-1}$ and 7 mg $= 7 \times 10^{-3}$ g: $$n = \frac{7 \times 10^{-3}}{14} = 5 \times 10^{-4}\ \mathrm{mol} \quad \Rightarrow \quad 5 \times 10^{-4} \times 6.022 \times 10^{23} = 3.011 \times 10^{20}\ \text{atoms}$$ Each $^{14}\mathrm{C}$ atom has $14 - 6 = 8$ neutrons (from the mass number, not the atomic number): $$N_n = 8 \times 3.011 \times 10^{20} = 2.409 \times 10^{21}\ \text{neutrons}$$ $$m_n = 2.409 \times 10^{21} \times 1.675 \times 10^{-27}\ \mathrm{kg} = 4.03 \times 10^{-6}\ \mathrm{kg}$$

(iii) Molar mass of $\mathrm{NH_3}$ is 17 g mol$^{-1}$: $$n = \frac{34 \times 10^{-3}}{17} = 2 \times 10^{-3}\ \mathrm{mol} \quad \Rightarrow \quad 1.204 \times 10^{21}\ \text{molecules}$$ Each molecule has $7 + 3 = 10$ protons (7 from N, 3 from H): $$N_p = 10 \times 1.204 \times 10^{21} = 1.204 \times 10^{22}\ \text{protons}$$ $$m_p = 1.204 \times 10^{22} \times 1.6726 \times 10^{-27}\ \mathrm{kg} = 2.01 \times 10^{-5}\ \mathrm{kg}$$

Temperature and pressure don't matter here. The proton count depends only on the mass of ammonia (so the moles), not on what volume it takes up. The "at STP" is just there to distract.

Ans: (i) $6.022 \times 10^{24}$ electrons. (ii) $2.409 \times 10^{21}$ neutrons, $4.03 \times 10^{-6}$ kg. (iii) $1.204 \times 10^{22}$ protons, $2.01 \times 10^{-5}$ kg; unchanged with temperature and pressure.

Watch out: Neutrons come from the mass number ($^{14}\mathrm{C}$ has 8, not 6). And T and P only matter if the sample is given as a volume of gas, not a mass.

Question 7: Counting electrons from a measured charge (Millikan)

(a) A certain particle carries $2.5 \times 10^{-16}$ C of static electric charge. Calculate the number of electrons present in it. (b) In Millikan's experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is $-1.282 \times 10^{-18}$ C, calculate the number of electrons present on it.

Answer:

Millikan's whole point was that charge comes in whole packets. Any charge on a drop is $q = ne$, where $n$ is an integer and $e = 1.602 \times 10^{-19}$ C. So I just divide: $n = q/e$.

(a) $$n = \frac{2.5 \times 10^{-16}\ \mathrm{C}}{1.602 \times 10^{-19}\ \mathrm{C}} = 1.56 \times 10^{3} = 1560\ \text{electrons}$$

(b) $$n = \frac{1.282 \times 10^{-18}\ \mathrm{C}}{1.602 \times 10^{-19}\ \mathrm{C}} = 8.0$$ So the drop has 8 extra electrons. The minus sign only tells me the drop gained electrons instead of losing them; it doesn't change the count.

Both answers came out as (nearly) whole numbers, which is exactly what should happen. If I'd got 8.4, that would mean I slipped in the arithmetic, not that there's a fraction of an electron.

Ans: (a) 1560 electrons. (b) 8 electrons.

Watch out: The charge on a drop can only be $\pm 1.602 \times 10^{-19}$ C, $\pm 3.204 \times 10^{-19}$ C, $\pm 4.806 \times 10^{-19}$ C and so on. If a list of charges is given and one is impossible, it's the one that isn't a whole-number multiple of $e$.

Question 8: Radio waves and the visible window

(a) The Vividh Bharati station of All India Radio, Delhi, broadcasts on a frequency of 1,368 kHz. Calculate the wavelength of the electromagnetic radiation emitted by the transmitter. Which part of the electromagnetic spectrum does it belong to? (b) The wavelength range of the visible spectrum extends from violet (400 nm) to red (750 nm). Express these wavelengths in frequencies (Hz).

Answer:

(a) Every electromagnetic wave travels at $c$ in vacuum, so $\lambda = c/\nu$. First I convert kHz to Hz: $1368\ \mathrm{kHz} = 1368 \times 10^{3}\ \mathrm{s^{-1}}$. $$\lambda = \frac{3.0 \times 10^{8}\ \mathrm{m\ s^{-1}}}{1368 \times 10^{3}\ \mathrm{s^{-1}}} = 219.3\ \mathrm{m}$$ A wavelength of a couple of hundred metres is a radio wave (the medium-wave AM band).

(b) Now I flip it to $\nu = c/\lambda$, with wavelengths in metres: $$\nu_{\text{violet}} = \frac{3.0 \times 10^{8}\ \mathrm{m\ s^{-1}}}{400 \times 10^{-9}\ \mathrm{m}} = 7.50 \times 10^{14}\ \mathrm{Hz}$$ $$\nu_{\text{red}} = \frac{3.0 \times 10^{8}\ \mathrm{m\ s^{-1}}}{750 \times 10^{-9}\ \mathrm{m}} = 4.00 \times 10^{14}\ \mathrm{Hz}$$ So visible light runs from $4.0 \times 10^{14}$ Hz (red) to $7.5 \times 10^{14}$ Hz (violet). Shorter wavelength, higher frequency.

Ans: (a) 219.3 m; radio waves. (b) Violet $7.5 \times 10^{14}$ Hz, red $4.0 \times 10^{14}$ Hz.

Watch out: $c = \nu\lambda$ is the easy part; the marks go on the unit conversions (kHz to Hz, nm to m). Worth remembering the visible window both ways: 400 to 750 nm, and $7.5$ down to $4.0 \times 10^{14}$ Hz, with violet at the short-wavelength, high-energy end.

Question 9: Yellow sodium light, and a wave known only by its period

(a) Yellow light emitted from a sodium lamp has a wavelength of 580 nm. Calculate the frequency and wavenumber of the yellow light. (b) Calculate the wavelength, frequency and wavenumber of a light wave whose period is $2.0 \times 10^{-10}$ s.

Answer:

(a) $\lambda = 580\ \mathrm{nm} = 580 \times 10^{-9}\ \mathrm{m} = 5.80 \times 10^{-7}$ m. Frequency first: $$\nu = \frac{c}{\lambda} = \frac{3.0 \times 10^{8}\ \mathrm{m\ s^{-1}}}{5.80 \times 10^{-7}\ \mathrm{m}} = 5.17 \times 10^{14}\ \mathrm{s^{-1}}$$ Wavenumber is just 1 over the wavelength, and it's usually quoted in $\mathrm{cm^{-1}}$. $580\ \mathrm{nm} = 5.80 \times 10^{-5}$ cm: $$\bar{\nu} = \frac{1}{\lambda} = \frac{1}{5.80 \times 10^{-7}\ \mathrm{m}} = 1.724 \times 10^{6}\ \mathrm{m^{-1}} = 1.724 \times 10^{4}\ \mathrm{cm^{-1}}$$ (Some versions of this problem say 5800 Å, which is the same 580 nm.)

(b) Here I only know the period, so I start from frequency. Frequency is oscillations per second, which is 1 over the period $T$: $$\nu = \frac{1}{T} = \frac{1}{2.0 \times 10^{-10}\ \mathrm{s}} = 5.0 \times 10^{9}\ \mathrm{s^{-1}}$$ Then wavelength: $$\lambda = \frac{c}{\nu} = \frac{3.0 \times 10^{8}\ \mathrm{m\ s^{-1}}}{5.0 \times 10^{9}\ \mathrm{s^{-1}}} = 6.0 \times 10^{-2}\ \mathrm{m} = 6.0\ \mathrm{cm}$$ And wavenumber: $$\bar{\nu} = \frac{1}{\lambda} = \frac{1}{6.0 \times 10^{-2}\ \mathrm{m}} = 16.7\ \mathrm{m^{-1}} = 0.167\ \mathrm{cm^{-1}}$$ A 6 cm wavelength at 5 GHz is a microwave.

Ans: (a) $\nu = 5.17 \times 10^{14}$ Hz, $\bar{\nu} = 1.724 \times 10^{6}\ \mathrm{m^{-1}} = 1.724 \times 10^{4}\ \mathrm{cm^{-1}}$. (b) $\nu = 5.0 \times 10^{9}$ Hz, $\lambda = 6.0 \times 10^{-2}$ m, $\bar{\nu} = 16.7\ \mathrm{m^{-1}}$.

Watch out: Wavenumber in $\mathrm{cm^{-1}}$ is 100 times smaller than in $\mathrm{m^{-1}}$, and I keep wanting to multiply instead of divide. Check with a 1 m wave: $\bar{\nu} = 1\ \mathrm{m^{-1}} = 0.01\ \mathrm{cm^{-1}}$.

Question 10: A neon sign — four questions about one wavelength

Neon gas is generally used in sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) the distance travelled by this radiation in 30 s, (c) the energy of one quantum, and (d) the number of quanta present if it produces 2 J of energy.

Answer: (a) First I put the wavelength in metres: $\lambda = 616\ \mathrm{nm} = 6.16 \times 10^{-7}$ m. Then $$\nu = \frac{c}{\lambda} = \frac{3.0 \times 10^{8}\ \mathrm{m\ s^{-1}}}{6.16 \times 10^{-7}\ \mathrm{m}} = 4.87 \times 10^{14}\ \mathrm{s^{-1}}$$

(b) This part has nothing to do with the wavelength. Light moves at $c$ whatever its colour, so I just multiply speed by time: $$d = c\,t = 3.0 \times 10^{8}\ \mathrm{m\ s^{-1}} \times 30\ \mathrm{s} = 9.0 \times 10^{9}\ \mathrm{m}$$

(c) One quantum carries $h\nu$: $$E = h\nu = 6.626 \times 10^{-34}\ \mathrm{J\ s} \times 4.87 \times 10^{14}\ \mathrm{s^{-1}} = 3.23 \times 10^{-19}\ \mathrm{J}$$ I checked it with the eV shortcut: $E = 1240/616 = 2.01$ eV $= 3.22 \times 10^{-19}$ J. Same thing.

(d) Total energy divided by energy per photon gives the count: $$N = \frac{E_{\text{total}}}{E_{\text{photon}}} = \frac{2\ \mathrm{J}}{3.23 \times 10^{-19}\ \mathrm{J}} = 6.2 \times 10^{18}\ \text{quanta}$$

Ans: (a) $4.87 \times 10^{14}$ Hz (b) $9.0 \times 10^{9}$ m (c) $3.23 \times 10^{-19}$ J (d) $6.2 \times 10^{18}$ quanta.

Watch out: In (d) keep the photon energy to three figures before dividing. If I'd rounded it to $3 \times 10^{-19}$ J the count would come out 8% off.

Question 11: Photon energies across the spectrum, and a mole of photons

(a) Find the energy of each of the photons which (i) correspond to light of frequency $3 \times 10^{15}$ Hz; (ii) have wavelength 0.50 Å. (b) Calculate the energy of one mole of photons of radiation whose frequency is $5 \times 10^{14}$ Hz. (c) The longest-wavelength doublet absorption transition of sodium is observed at 589 and 589.6 nm. Calculate the frequency of each transition and the energy difference between the two excited states.

Answer: (a)(i) I'm given the frequency, so $E = h\nu$ straight away: $$E = 6.626 \times 10^{-34}\ \mathrm{J\ s} \times 3 \times 10^{15}\ \mathrm{s^{-1}} = 1.99 \times 10^{-18}\ \mathrm{J}$$

(a)(ii) This time I'm given the wavelength, so $E = hc/\lambda$, with $0.50\ \text{\AA} = 0.50 \times 10^{-10}$ m: $$E = \frac{6.626 \times 10^{-34}\ \mathrm{J\ s} \times 3.0 \times 10^{8}\ \mathrm{m\ s^{-1}}}{0.50 \times 10^{-10}\ \mathrm{m}} = 3.98 \times 10^{-15}\ \mathrm{J}$$ That's about 25 keV, an X-ray photon, roughly two thousand times more energetic than the ultraviolet photon in (i).

(b) I work out one photon first, then multiply by Avogadro's number: $$E_{\text{photon}} = 6.626 \times 10^{-34} \times 5 \times 10^{14} = 3.313 \times 10^{-19}\ \mathrm{J}$$ $$E_{\text{mole}} = 3.313 \times 10^{-19}\ \mathrm{J} \times 6.022 \times 10^{23}\ \mathrm{mol^{-1}} = 1.995 \times 10^{5}\ \mathrm{J\ mol^{-1}} = 199.5\ \mathrm{kJ\ mol^{-1}}$$

(c) Two wavelengths, so two frequencies: $$\nu_1 = \frac{3.0 \times 10^{8}}{589 \times 10^{-9}} = 5.093 \times 10^{14}\ \mathrm{Hz}, \qquad \nu_2 = \frac{3.0 \times 10^{8}}{589.6 \times 10^{-9}} = 5.088 \times 10^{14}\ \mathrm{Hz}$$ The energy gap between the two excited states is just the difference of the two photon energies: $$\Delta E = h(\nu_1 - \nu_2) = 6.626 \times 10^{-34} \times (5.093 - 5.088) \times 10^{14} = 3.3 \times 10^{-22}\ \mathrm{J}$$ If I keep more digits, $\nu_1 - \nu_2 = 5.18 \times 10^{11}$ Hz and $\Delta E = 3.43 \times 10^{-22}$ J, about $2.1 \times 10^{-3}$ eV. It's a tiny splitting, which is why the two lines sit so close together.

Ans: (a) (i) $1.99 \times 10^{-18}$ J (ii) $3.98 \times 10^{-15}$ J. (b) $199.5$ kJ mol$^{-1}$. (c) $5.093 \times 10^{14}$ Hz and $5.088 \times 10^{14}$ Hz; $\Delta E \approx 3.4 \times 10^{-22}$ J.

Watch out: In (c) I'm subtracting two nearly equal numbers, so rounding the frequencies early wrecks the answer; I carry four figures. A cleaner route is $\Delta E = hc\,\Delta\lambda/\lambda^2$, which gives $1240 \times 0.6/589^2 = 2.1 \times 10^{-3}$ eV in one line with no subtraction.

Question 12: Photons per second from a bulb

(a) A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb. (b) A 25 watt bulb emits monochromatic yellow light of wavelength 0.57 $\mu$m. Calculate the rate of emission of quanta per second.

Answer: A watt is a joule per second, so 100 W means the bulb gives out $100\ \mathrm{J\ s^{-1}}$. Photons per second is just energy per second divided by energy per photon.

(a) Energy of one 400 nm photon: $$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{400 \times 10^{-9}} = 4.969 \times 10^{-19}\ \mathrm{J}$$ $$\text{Rate} = \frac{100\ \mathrm{J\ s^{-1}}}{4.969 \times 10^{-19}\ \mathrm{J}} = 2.012 \times 10^{20}\ \text{photons per second}$$

(b) Same idea, with $0.57\ \mu\mathrm{m} = 0.57 \times 10^{-6}$ m: $$E = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{0.57 \times 10^{-6}} = 3.487 \times 10^{-19}\ \mathrm{J}$$ $$\text{Rate} = \frac{25\ \mathrm{J\ s^{-1}}}{3.487 \times 10^{-19}\ \mathrm{J}} = 7.17 \times 10^{19}\ \text{quanta per second}$$

Quick check: the rate goes as $P\lambda$. Bulb (b) has a quarter of the power but $1.425$ times the wavelength, so it should give $0.25 \times 1.425 = 0.356$ times as many photons. $0.356 \times 2.012 \times 10^{20} = 7.17 \times 10^{19}$, which matches.

Ans: (a) $2.012 \times 10^{20}$ photons s$^{-1}$. (b) $7.17 \times 10^{19}$ quanta s$^{-1}$.

Watch out: $N/t = P\lambda/hc$. If a question says only 10% of the power goes into light, use $0.1P$ on top. And check the wavelength unit before anything else, since $\mu$m and Å both turn up here.

Question 13: Counting photons — a joule of X-rays, a starlight detector, a laser and a pulse

(a) What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy? (b) A photon detector receives a total of $3.15 \times 10^{-18}$ J from radiation of 600 nm. Calculate the number of photons received. (c) A nitrogen laser produces radiation at 337.1 nm. If the number of photons emitted is $5.6 \times 10^{24}$, calculate the power of this laser. (d) A pulsed radiation source has a duration of 2 ns and emits $2.5 \times 10^{15}$ photons during the pulse. Calculate the energy of the source.

Answer: (a) First the unit: $4000\ \mathrm{pm} = 4000 \times 10^{-12}\ \mathrm{m} = 4.0 \times 10^{-9}$ m, which is 4 nm, a soft X-ray. Then one photon's energy and the count: $$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{4.0 \times 10^{-9}} = 4.97 \times 10^{-17}\ \mathrm{J}$$ $$N = \frac{1\ \mathrm{J}}{4.97 \times 10^{-17}\ \mathrm{J}} = 2.01 \times 10^{16}\ \text{photons}$$

(b) $600\ \mathrm{nm} = 6.0 \times 10^{-7}$ m: $$E = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{6.0 \times 10^{-7}} = 3.313 \times 10^{-19}\ \mathrm{J}$$ $$N = \frac{3.15 \times 10^{-18}}{3.313 \times 10^{-19}} = 9.5 \approx 10\ \text{photons}$$ You can't have half a photon, so I round it to about 10. That's how faint starlight is.

(c) Energy of one 337.1 nm photon: $$E = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{337.1 \times 10^{-9}} = 5.897 \times 10^{-19}\ \mathrm{J}$$ Total energy of $5.6 \times 10^{24}$ photons $= 5.6 \times 10^{24} \times 5.897 \times 10^{-19} = 3.30 \times 10^{6}$ J. The question calls this the power, so it's treating the photon count as per second: $$P = 3.3 \times 10^{6}\ \mathrm{J\ s^{-1}} = 3.3 \times 10^{6}\ \mathrm{W}$$

(d) The usual way of reading this one is to take the pulse duration as the period of the radiation, so $\nu = 1/T = 1/(2 \times 10^{-9}\ \mathrm{s}) = 5.0 \times 10^{8}$ Hz: $$E_{\text{photon}} = h\nu = 6.626 \times 10^{-34} \times 5.0 \times 10^{8} = 3.31 \times 10^{-25}\ \mathrm{J}$$ $$E_{\text{source}} = 2.5 \times 10^{15} \times 3.31 \times 10^{-25} = 8.28 \times 10^{-10}\ \mathrm{J}$$ Strictly, how long a pulse lasts and the frequency of its light are two different things, but this is the convention the answer key uses, and the arithmetic is the point.

Ans: (a) $2.01 \times 10^{16}$ photons (b) about 10 photons (c) $3.3 \times 10^{6}$ W (d) $8.28 \times 10^{-10}$ J.

Watch out: A count like 9.5 photons doesn't mean anything on its own. Round a photon count to a whole number and say you did.

Question 14: Threshold frequency and work function

(a) The threshold frequency $\nu_0$ for a metal is $7.0 \times 10^{14}\ \mathrm{s^{-1}}$. Calculate the kinetic energy of an electron emitted when radiation of frequency $\nu = 1.0 \times 10^{15}\ \mathrm{s^{-1}}$ hits the metal. (b) Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate the threshold frequency and the work function of the metal.

Answer: Einstein's equation is $h\nu = W_0 + \tfrac{1}{2}m_e v^2$, with $W_0 = h\nu_0$. So the kinetic energy is whatever's left over: $KE = h(\nu - \nu_0)$.

(a) $$KE = 6.626 \times 10^{-34}\ \mathrm{J\ s} \times (10.0 \times 10^{14} - 7.0 \times 10^{14})\ \mathrm{s^{-1}} = 6.626 \times 10^{-34} \times 3.0 \times 10^{14} = 1.988 \times 10^{-19}\ \mathrm{J}$$ That's about 1.24 eV.

(b) "Zero velocity" tells me the light is exactly at threshold, so $\lambda_0 = 6800\ \text{\AA} = 6.8 \times 10^{-7}$ m and $$\nu_0 = \frac{c}{\lambda_0} = \frac{3.0 \times 10^{8}}{6.8 \times 10^{-7}} = 4.41 \times 10^{14}\ \mathrm{s^{-1}}$$ The work function is then $$W_0 = h\nu_0 = 6.626 \times 10^{-34} \times 4.41 \times 10^{14} = 2.92 \times 10^{-19}\ \mathrm{J} = \frac{2.92 \times 10^{-19}}{1.602 \times 10^{-19}} = 1.82\ \mathrm{eV}$$ Quick check with the shortcut: $W_0 = 1240/680 = 1.82$ eV.

Ans: (a) $1.988 \times 10^{-19}$ J. (b) $\nu_0 = 4.41 \times 10^{14}$ Hz, $W_0 = 2.92 \times 10^{-19}$ J $= 1.82$ eV.

Watch out: "Emitted with zero kinetic energy" or "just sufficient to eject" means threshold: $\lambda = \lambda_0$, $\nu = \nu_0$, and $h\nu_0$ is the work function. Also, $KE_{\max}$ depends on frequency only; brighter light just gives more electrons per second, not faster ones.

Question 15: Photon energy, kinetic energy and photoelectron velocity

(a) A photon of wavelength $4 \times 10^{-7}$ m strikes a metal surface whose work function is 2.13 eV. Calculate (i) the energy of the photon in eV, (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron. (1 eV $= 1.6020 \times 10^{-19}$ J.) (b) When radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of $1.68 \times 10^{5}\ \mathrm{J\ mol^{-1}}$. What is the minimum energy needed to remove an electron from sodium, and what is the maximum wavelength that will cause a photoelectron to be emitted?

Answer: (a)(i) Photon energy, then converted to eV: $$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{4 \times 10^{-7}} = 4.97 \times 10^{-19}\ \mathrm{J} = \frac{4.97 \times 10^{-19}}{1.602 \times 10^{-19}} = 3.10\ \mathrm{eV}$$

(a)(ii) The kinetic energy is what's left after paying the work function: $$KE = 3.10 - 2.13 = 0.97\ \mathrm{eV} = 0.97 \times 1.602 \times 10^{-19} = 1.55 \times 10^{-19}\ \mathrm{J}$$

(a)(iii) For the speed I use $KE = \tfrac{1}{2}m_e v^2$, and here the kinetic energy has to be in joules: $$v = \sqrt{\frac{2\,KE}{m_e}} = \sqrt{\frac{2 \times 1.55 \times 10^{-19}\ \mathrm{J}}{9.1 \times 10^{-31}\ \mathrm{kg}}} = \sqrt{3.41 \times 10^{11}}\ \mathrm{m\ s^{-1}} = 5.84 \times 10^{5}\ \mathrm{m\ s^{-1}}$$

(b) The kinetic energy is given per mole, so I first bring it down to one electron: $$KE = \frac{1.68 \times 10^{5}\ \mathrm{J\ mol^{-1}}}{6.022 \times 10^{23}\ \mathrm{mol^{-1}}} = 2.79 \times 10^{-19}\ \mathrm{J}$$ Energy of the 300 nm photon: $$E = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{300 \times 10^{-9}} = 6.626 \times 10^{-19}\ \mathrm{J}$$ The minimum energy to remove an electron is the work function, photon energy minus kinetic energy: $$W_0 = E - KE = 6.626 \times 10^{-19} - 2.79 \times 10^{-19} = 3.84 \times 10^{-19}\ \mathrm{J}$$ The maximum wavelength is the threshold wavelength, where the whole photon goes into the work function and nothing is left for motion: $$\lambda_0 = \frac{hc}{W_0} = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{3.84 \times 10^{-19}} = 5.17 \times 10^{-7}\ \mathrm{m} = 517\ \mathrm{nm}$$ That's green light, so sodium responds to anything bluer than 517 nm.

Ans: (a) (i) 3.10 eV (ii) 0.97 eV $= 1.55 \times 10^{-19}$ J (iii) $5.84 \times 10^{5}\ \mathrm{m\ s^{-1}}$. (b) $W_0 = 3.84 \times 10^{-19}$ J; $\lambda_0 = 517$ nm.

Watch out: I do the energy subtraction in eV ($3.10 - 2.13$) but the velocity in joules and kilograms; mixing the two is the classic slip. A handy scale: 1 eV of kinetic energy gives an electron $5.93 \times 10^{5}\ \mathrm{m\ s^{-1}}$, so 0.97 eV gives $\sqrt{0.97} \times 5.93 \times 10^{5} = 5.84 \times 10^{5}\ \mathrm{m\ s^{-1}}$ straight away.

Question 16: Ionisation energy of sodium, and caesium under 500 nm light

(a) Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol$^{-1}$. (b) The work function for caesium is 1.9 eV. Calculate (i) the threshold wavelength and (ii) the threshold frequency of the radiation. If caesium is irradiated with light of wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

Answer: (a) "Just sufficient" means one photon carries exactly the ionisation energy, no more. So I find the photon energy: $$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{242 \times 10^{-9}} = 8.21 \times 10^{-19}\ \mathrm{J\ atom^{-1}}$$ Then scale it up to a mole: $$IE = 8.21 \times 10^{-19}\ \mathrm{J} \times 6.022 \times 10^{23}\ \mathrm{mol^{-1}} = 4.945 \times 10^{5}\ \mathrm{J\ mol^{-1}} = 494.5\ \mathrm{kJ\ mol^{-1}}$$ Shortcut check: $1240/242 = 5.12$ eV, and $1\ \mathrm{eV\ atom^{-1}} = 96.5\ \mathrm{kJ\ mol^{-1}}$, so $5.12 \times 96.5 = 494$ kJ mol$^{-1}$.

(b)(i) I convert the work function to joules first: $W_0 = 1.9\ \mathrm{eV} = 1.9 \times 1.602 \times 10^{-19} = 3.04 \times 10^{-19}$ J. Then $$\lambda_0 = \frac{hc}{W_0} = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{3.04 \times 10^{-19}} = 6.53 \times 10^{-7}\ \mathrm{m} = 653\ \mathrm{nm}$$

(b)(ii) Threshold frequency: $$\nu_0 = \frac{W_0}{h} = \frac{3.04 \times 10^{-19}}{6.626 \times 10^{-34}} = 4.59 \times 10^{14}\ \mathrm{s^{-1}}$$

For the 500 nm light, the photon energy is $1240/500 = 2.48$ eV, so the leftover kinetic energy is $$KE = 2.48 - 1.9 = 0.58\ \mathrm{eV} = 0.58 \times 1.602 \times 10^{-19} = 9.3 \times 10^{-20}\ \mathrm{J}$$ and the speed follows from $\tfrac{1}{2}m_e v^2 = KE$: $$v = \sqrt{\frac{2 \times 9.3 \times 10^{-20}}{9.1 \times 10^{-31}}} = \sqrt{2.04 \times 10^{11}} = 4.5 \times 10^{5}\ \mathrm{m\ s^{-1}}$$

Ans: (a) $494.5$ kJ mol$^{-1}$. (b) $\lambda_0 = 653$ nm, $\nu_0 = 4.59 \times 10^{14}$ Hz; at 500 nm, $KE = 0.58$ eV $= 9.3 \times 10^{-20}$ J and $v = 4.5 \times 10^{5}\ \mathrm{m\ s^{-1}}$.

Watch out: Ionisation energy from a threshold wavelength is just the photoelectric effect on a gas atom; $1\ \mathrm{eV\ atom^{-1}} = 96.5\ \mathrm{kJ\ mol^{-1}}$ converts it to the per-mole unit. Caesium's 1.9 eV is the lowest work function of the common metals, which is why even red light works on it and why it's used in photocells.

Question 17: Planck's constant from photoelectric data

Sodium metal is irradiated with three wavelengths, and the maximum speed of the photoelectrons is measured each time:

$\lambda$ (nm)	500	450	400
$v$ ($\times 10^{5}\ \mathrm{m\ s^{-1}}$)	2.55	4.35	5.35

Calculate (a) the threshold wavelength and (b) Planck's constant.

Answer: The plan: Einstein's equation $\tfrac{1}{2}m_e v^2 = h\nu - h\nu_0$ is a straight line of kinetic energy against frequency, slope $h$ and intercept $-h\nu_0$. Two points fix the slope; the third is a check. The speeds are in multiples of $10^{5}\ \mathrm{m\ s^{-1}}$; with any smaller unit the kinetic energies would be far below a photon's energy and the data wouldn't make sense.

Frequencies first: $$\nu_1 = \frac{3.0 \times 10^{8}}{500 \times 10^{-9}} = 6.00 \times 10^{14}\ \mathrm{Hz}, \quad \nu_2 = 6.67 \times 10^{14}\ \mathrm{Hz}, \quad \nu_3 = 7.50 \times 10^{14}\ \mathrm{Hz}$$

Kinetic energies with $m_e = 9.1 \times 10^{-31}$ kg: $$KE_1 = \tfrac{1}{2}(9.1 \times 10^{-31})(2.55 \times 10^{5})^2 = 2.96 \times 10^{-20}\ \mathrm{J}$$ $$KE_2 = \tfrac{1}{2}(9.1 \times 10^{-31})(4.35 \times 10^{5})^2 = 8.61 \times 10^{-20}\ \mathrm{J}$$ $$KE_3 = \tfrac{1}{2}(9.1 \times 10^{-31})(5.35 \times 10^{5})^2 = 1.30 \times 10^{-19}\ \mathrm{J}$$

(b) I take the slope between the two extreme points, since that's the widest span: $$h = \frac{KE_3 - KE_1}{\nu_3 - \nu_1} = \frac{1.30 \times 10^{-19} - 0.296 \times 10^{-19}}{7.50 \times 10^{14} - 6.00 \times 10^{14}} = \frac{1.006 \times 10^{-19}}{1.50 \times 10^{14}} = 6.7 \times 10^{-34}\ \mathrm{J\ s}$$ Very close to the accepted $6.626 \times 10^{-34}$ J s. The middle point gives slopes of $8.5$ and $5.3 \times 10^{-34}$ with its neighbours, which is just scatter in real data, and that's why I use the widest span or a best-fit line.

(a) Threshold frequency from point 1 and this $h$: $$\nu_0 = \nu_1 - \frac{KE_1}{h} = 6.00 \times 10^{14} - \frac{2.96 \times 10^{-20}}{6.7 \times 10^{-34}} = 6.00 \times 10^{14} - 0.44 \times 10^{14} = 5.56 \times 10^{14}\ \mathrm{Hz}$$ $$\lambda_0 = \frac{c}{\nu_0} = \frac{3.0 \times 10^{8}}{5.56 \times 10^{14}} = 5.4 \times 10^{-7}\ \mathrm{m} = 540\ \mathrm{nm}$$ The answer key usually quotes about 531 nm with $h = 6.66 \times 10^{-34}$ J s. The spread comes from the data, not from a mistake.

Ans: (a) $\lambda_0 \approx 540$ nm (about 530 to 540 nm depending on the fit). (b) $h \approx 6.7 \times 10^{-34}$ J s.

Watch out: On a $KE_{\max}$ against $\nu$ graph, the slope is $h$ for every metal; only the intercept ($-W_0$, or $\nu_0$ on the $x$-axis) changes with the work function.

Question 18: Stopping potential, and a tightly bound inner electron

(a) The ejection of photoelectrons from silver can be stopped by applying a voltage of 0.35 V when radiation of 256.7 nm is used. Calculate the work function of silver. (b) A photon of wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected with a velocity of $1.5 \times 10^{7}\ \mathrm{m\ s^{-1}}$. Calculate the energy with which it is bound to the nucleus.

Answer: (a) If a reverse voltage $V_s$ just stops the fastest electrons, then $KE_{\max} = eV_s$. With $V_s = 0.35$ V, that's $KE_{\max} = 0.35$ eV, no conversion needed. That's the nice thing about working in eV.

Photon energy: $$E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{256.7 \times 10^{-9}} = 7.74 \times 10^{-19}\ \mathrm{J} = \frac{7.74 \times 10^{-19}}{1.602 \times 10^{-19}} = 4.83\ \mathrm{eV}$$ Work function: $$W_0 = E - KE_{\max} = 4.83 - 0.35 = 4.48\ \mathrm{eV} = 7.18 \times 10^{-19}\ \mathrm{J}$$

(b) A 150 pm photon is a hard X-ray. Its energy: $$E = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{150 \times 10^{-12}} = 1.325 \times 10^{-15}\ \mathrm{J}$$ The electron carries away $$KE = \tfrac{1}{2}m_e v^2 = \tfrac{1}{2} \times 9.1 \times 10^{-31} \times (1.5 \times 10^{7})^2 = 1.02 \times 10^{-16}\ \mathrm{J}$$ The rest of the photon's energy went into freeing the electron, so that's the binding energy: $$BE = 1.325 \times 10^{-15} - 0.102 \times 10^{-15} = 1.22 \times 10^{-15}\ \mathrm{J} = \frac{1.22 \times 10^{-15}}{1.602 \times 10^{-19}} = 7.6 \times 10^{3}\ \mathrm{eV}$$

Ans: (a) $W_0 = 4.48$ eV $= 7.18 \times 10^{-19}$ J. (b) Binding energy $1.22 \times 10^{-15}$ J, about 7.6 keV.

Watch out: $eV_s = h\nu - W_0$: a stopping potential in volts is the kinetic energy in eV directly. Part (b) is the same equation with "binding energy" in place of the work function. Before subtracting in (b), I wrote both energies with the same power of ten ($1.325$ and $0.102 \times 10^{-15}$ J); that's where the usual slip happens.