Dual Nature of Matter and Radiation — Easy Practice (Set 1)

Quick Recap — Photoelectric Effect

• Dual nature: light behaves both as a wave (interference, diffraction) and as a particle (the photoelectric effect).
• Photon energy $E = h\nu = \dfrac{hc}{\lambda}$; a handy form is $E(\text{eV}) = \dfrac{1240}{\lambda(\text{nm})}$.
• Einstein's equation: $KE_{max} = h\nu - \phi$, where $\phi = h\nu_0$ is the work function.
• Intensity sets the number of photoelectrons (the photocurrent); frequency sets their maximum kinetic energy.
• Stopping potential: $eV_0 = KE_{max}$; it depends on the frequency, not the intensity.

Beyond-NCERT JEE Formulae

A compact working sheet for fast problem-solving. NCERT supplies the physics; collected here are the shortcuts, sign conventions and graph facts that JEE questions actually reward.

1. Photoelectric Effect

One photon is absorbed by one electron. Its energy $h\nu$ pays the work function $\phi$ first, and whatever remains appears as kinetic energy:

$$h\nu = \phi + KE_{max}$$

Hence $KE_{max} = h\nu - \phi = h(\nu - \nu_0) = eV_0$, where $V_0$ is the stopping potential and $\nu_0 = \dfrac{\phi}{h}$ is the threshold frequency.

• Threshold wavelength: $\lambda_0 = \dfrac{hc}{\phi}$. Emission occurs only for $\lambda \le \lambda_0$ (equivalently $\nu \ge \nu_0$), no matter how intense the beam.
• Stopping potential from wavelength: $eV_0 = \dfrac{hc}{\lambda} - \phi$, so numerically the value of $V_0$ in volts equals $KE_{max}$ in eV.

When to use. Any "light of wavelength $\lambda$ on a metal of work function $\phi$" problem: find $KE_{max} = \dfrac{hc}{\lambda} - \phi$, then read off $V_0 = \dfrac{KE_{max}}{e}$.

[JEE Tip] A graph of $V_0$ against frequency is a straight line $V_0 = \dfrac{h}{e}\,\nu - \dfrac{\phi}{e}$. Its slope is $\dfrac{h}{e}$ and is identical for every metal (this is exactly how $h$ is measured); the frequency-intercept gives $\nu_0$, and the vertical-intercept is $-\dfrac{\phi}{e}$. Two different metals therefore give parallel lines.

2. de Broglie Wavelength of Matter

Every particle of momentum $p$ has a wavelength

$$\lambda = \dfrac{h}{p} = \dfrac{h}{mv} = \dfrac{h}{\sqrt{2mK}} = \dfrac{h}{\sqrt{2mqV}}$$

• From kinetic energy $K$: use $\lambda = \dfrac{h}{\sqrt{2mK}}$ (a thermal particle has $K = \dfrac{3}{2}k_BT$).
• From accelerating voltage $V$: a charge $q$ gains $K = qV$, giving $\lambda = \dfrac{h}{\sqrt{2mqV}}$.
• Electron shortcut: $\lambda = \dfrac{12.27}{\sqrt{V}}$ angstrom, with $V$ in volts. This single number cracks most electron problems in seconds; for instance $150$ V gives almost exactly $1$ angstrom.

When to use. The phrase "accelerated through $V$ volts" points straight at the electron shortcut; a stated "kinetic energy $K$" or "temperature $T$" points at the $\sqrt{2mK}$ form instead.

[JEE Tip] Scaling laws settle most of these questions: for equal $K$ the rule is $\lambda \propto \dfrac{1}{\sqrt{m}}$; for equal $V$ it is $\lambda \propto \dfrac{1}{\sqrt{mq}}$; and for equal momentum the wavelengths are simply equal. A handy proton companion is $\lambda = \dfrac{0.286}{\sqrt{V}}$ angstrom. Above a few keV an electron needs the relativistic form $\lambda = \dfrac{h}{\sqrt{2mK\left(1 + \dfrac{K}{2mc^2}\right)}}$.

3. Photon: Energy and Momentum

A photon is a massless quantum travelling at $c$, carrying both energy and momentum:

$$E = h\nu = \dfrac{hc}{\lambda}, \qquad p = \dfrac{h}{\lambda} = \dfrac{E}{c}$$

• Photons per second from a source of power $P$: divide the power by the energy of one photon, $N = \dfrac{P}{E} = \dfrac{P\lambda}{hc}$.
• Photon flux from an intensity $I$ falling on unit area: the number arriving per unit area each second is $\dfrac{I\lambda}{hc}$.
• Radiation force on a target: $F = \dfrac{P}{c}$ when the light is fully absorbed, and $F = \dfrac{2P}{c}$ when it is fully reflected.

When to use. "A laser of power $P$ at $\lambda$ nm" almost always wants $N = \dfrac{P\lambda}{hc}$; a black or absorbing surface wants $F = \dfrac{P}{c}$, while a mirror wants $\dfrac{2P}{c}$.

[JEE Tip] In $p = \dfrac{E}{c}$ the energy must be in joules, never electronvolts. Forgetting to multiply the eV figure by $1.6 \times 10^{-19}$ before dividing by $c$ is the single commonest slip in this chapter.

4. The Constant That Saves Time

Rather than juggling $h$, $c$ and stray powers of ten, memorise

$$hc = 1240 \text{ eV nm} \quad\Longrightarrow\quad E\text{(eV)} = \dfrac{1240}{\lambda\text{(nm)}}$$

When to use. Every photon-energy or stopping-potential problem in which the wavelength is quoted in nm and the answer is wanted in eV or volts.

[JEE Tip] Two more shortcuts flow from the same constant: the threshold wavelength $\lambda_0\text{(nm)} = \dfrac{1240}{\phi\text{(eV)}}$, and the X-ray (Duane-Hunt) cut-off $\lambda_{min}\text{(pm)} = \dfrac{1240}{V\text{(kV)}}$. Each lets you jump straight from an energy to a wavelength with no unit conversions at all.

Solved Examples — Beyond-NCERT Formulae

Constants used throughout: $hc = 1240$ eV nm, $h = 6.63 \times 10^{-34}$ J s, $c = 3 \times 10^8$ m/s, $1$ eV $= 1.6 \times 10^{-19}$ J, $m_e = 9.11 \times 10^{-31}$ kg, $m_p = 1.67 \times 10^{-27}$ kg, and $e = 1.6 \times 10^{-19}$ C.

Example 1 — Stopping potential from wavelength and work function

Problem. Light of wavelength $248$ nm falls on a metal of work function $2.0$ eV. Find the maximum kinetic energy of the photoelectrons and the stopping potential.

Formula. $KE_{max} = \dfrac{hc}{\lambda} - \phi$, and the stopping potential follows from $eV_0 = KE_{max}$.

Solution. The photon energy is $\dfrac{hc}{\lambda} = \dfrac{1240}{248} = 5.0$ eV. Subtracting the work function, $KE_{max} = 5.0 - 2.0 = 3.0$ eV. Since $eV_0 = KE_{max}$, the stopping potential is $V_0 = 3.0$ V.

Answer. $KE_{max} = 3.0$ eV and $V_0 = 3.0$ V.

Example 2 — Threshold wavelength and threshold frequency

Problem. A metal has a work function of $3.1$ eV. Find its threshold wavelength and threshold frequency.

Formula. $\lambda_0 = \dfrac{hc}{\phi}$ and $\nu_0 = \dfrac{c}{\lambda_0}$.

Solution. The threshold wavelength is $\lambda_0 = \dfrac{1240}{3.1} = 400$ nm. Writing this as $4.0 \times 10^{-7}$ m, the threshold frequency is $\nu_0 = \dfrac{3 \times 10^8}{4.0 \times 10^{-7}} = 7.5 \times 10^{14}$ Hz.

Answer. $\lambda_0 = 400$ nm and $\nu_0 = 7.5 \times 10^{14}$ Hz. Any light of longer wavelength than $400$ nm fails to eject electrons, however intense it is.

Example 3 — de Broglie wavelength of an accelerated electron

Problem. An electron starts from rest and is accelerated through $150$ V. Find its de Broglie wavelength.

Formula. Electron shortcut $\lambda = \dfrac{12.27}{\sqrt{V}}$ angstrom, cross-checked against $\lambda = \dfrac{h}{\sqrt{2m_e eV}}$.

Solution. The shortcut gives $\lambda = \dfrac{12.27}{\sqrt{150}} = \dfrac{12.27}{12.25} = 1.00$ angstrom. As a full check, $\sqrt{2m_e eV} = \sqrt{2(9.11 \times 10^{-31})(1.6 \times 10^{-19})(150)} = 6.61 \times 10^{-24}$ kg m/s, so $\lambda = \dfrac{6.63 \times 10^{-34}}{6.61 \times 10^{-24}} = 1.00 \times 10^{-10}$ m.

Answer. $\lambda \approx 1.00$ angstrom, that is about $100$ pm — the classic result that a $150$ V electron has a wavelength near one angstrom.

Example 4 — de Broglie wavelength of an accelerated proton

Problem. A proton starts from rest and is accelerated through $1000$ V. Find its de Broglie wavelength.

Formula. General charged-particle form $\lambda = \dfrac{h}{\sqrt{2mqV}}$, with the proton shortcut $\lambda = \dfrac{0.286}{\sqrt{V}}$ angstrom as a check.

Solution. With $m = 1.67 \times 10^{-27}$ kg and $q = 1.6 \times 10^{-19}$ C, the momentum is $\sqrt{2mqV} = \sqrt{2(1.67 \times 10^{-27})(1.6 \times 10^{-19})(1000)} = 7.31 \times 10^{-22}$ kg m/s, so $\lambda = \dfrac{6.63 \times 10^{-34}}{7.31 \times 10^{-22}} = 9.07 \times 10^{-13}$ m. The shortcut agrees: $\dfrac{0.286}{\sqrt{1000}} = 0.00904$ angstrom.

Answer. $\lambda \approx 0.91$ pm. Because the proton is far heavier than an electron, at the same accelerating voltage its wavelength is about $43$ times shorter.

Example 5 — Photons per second from a laser

Problem. A laser emits $2.0$ mW of light at a wavelength of $400$ nm. How many photons does it emit each second?

Formula. $N = \dfrac{P}{E} = \dfrac{P\lambda}{hc}$, where $E$ is the energy of a single photon.

Solution. The energy of one photon is $E = \dfrac{1240}{400} = 3.1$ eV, which equals $3.1 \times 1.6 \times 10^{-19} = 4.96 \times 10^{-19}$ J. Dividing the power (in watts) by this, $N = \dfrac{2.0 \times 10^{-3}}{4.96 \times 10^{-19}} = 4.0 \times 10^{15}$ per second.

Answer. About $4.0 \times 10^{15}$ photons per second. The power must be converted from mW to W before dividing.

Example 6 — Momentum of a photon

Problem. Find the momentum of a photon of red light of wavelength $660$ nm.

Formula. $p = \dfrac{h}{\lambda}$, cross-checked with $p = \dfrac{E}{c}$.

Solution. Directly, $p = \dfrac{6.63 \times 10^{-34}}{660 \times 10^{-9}} = 1.00 \times 10^{-27}$ kg m/s. As a check through the energy, $E = \dfrac{1240}{660} = 1.88$ eV, which is $3.01 \times 10^{-19}$ J, giving $\dfrac{E}{c} = \dfrac{3.01 \times 10^{-19}}{3 \times 10^8} = 1.00 \times 10^{-27}$ kg m/s.

Answer. $p \approx 1.0 \times 10^{-27}$ kg m/s. The two routes agree because $E = pc$ for a photon.

Example 7 — Radiation force of a laser beam

Problem. A $30$ W laser beam strikes a target. Find the force it exerts when the target (a) fully absorbs the light and (b) fully reflects it.

Formula. Momentum is delivered at the rate $\dfrac{P}{c}$; a fully absorbed beam gives $F = \dfrac{P}{c}$, while a fully reflected beam gives $F = \dfrac{2P}{c}$.

Solution. (a) For full absorption, $F = \dfrac{P}{c} = \dfrac{30}{3 \times 10^8} = 1.0 \times 10^{-7}$ N. (b) A mirror reverses each photon's momentum, doubling the impulse, so $F = \dfrac{2P}{c} = 2.0 \times 10^{-7}$ N.

Answer. $1.0 \times 10^{-7}$ N when absorbed and $2.0 \times 10^{-7}$ N when reflected. The force is minute, yet this same radiation pressure is what propels a solar sail.