JEE Physics Dual Nature Of Radiation And Matter Notes Set 05

Photoelectric Effect and Einstein's Equation

When light of sufficiently high frequency falls on a metal, electrons are emitted almost instantly. Light behaves as packets of energy (photons), each of energy \( E=hf=\frac{hc}{\lambda} \) and momentum \( p=\frac{E}{c}=\frac h\lambda \). A single photon gives all its energy to a single electron. Part of the energy, the work function \( \phi \), is spent in escaping from the metal and the rest appears as kinetic energy, so \( hf=\phi+K_{max} \). The threshold frequency is \( f_0=\phi/h \) and the threshold wavelength is \( \lambda_0=hc/\phi \). A useful value is \( hc=1240 \) eV nm.
The stopping potential \( V_0 \) satisfies \( eV_0=K_{max}=hf-\phi \). A graph of \( V_0 \) against f is a straight line of slope \( h/e \), the same for all metals, whose intercept on the frequency axis is the threshold frequency.

Effect of Intensity, Frequency and Distance

The number of photons per second, and so the saturation photocurrent, is proportional to the intensity (for light above threshold). The maximum kinetic energy and the stopping potential depend only on frequency and the metal, not on intensity. Moving a point source to double the distance reduces the intensity to one quarter, which cuts the current to one quarter but leaves the stopping potential unchanged.
If the frequency is doubled, the new kinetic energy is \( 2hf-\phi=2K+\phi \), which is more than double the old value. Increasing the work function with the same photons lowers the electron energy but not the number of electrons, provided the photon energy remains above threshold.

De Broglie Waves

Moving matter has an associated wavelength \( \lambda=\frac hp=\frac{h}{mv} \). For a particle of kinetic energy K, \( \lambda=\frac{h}{\sqrt{2mK}} \), and for a charge q accelerated through potential V, \( \lambda=\frac{h}{\sqrt{2mqV}} \). For an electron, \( \lambda\approx\frac{12.27}{\sqrt V} \) Å. At equal kinetic energy the lighter particle has the longer wavelength; at equal momentum the wavelengths are equal. For gas molecules at temperature T, \( \lambda\propto T^{-1/2} \). A particle confined in a box of length a, with nodes at both ends, has \( \lambda=\frac{2a}{n} \) and energy \( E_n=\frac{n^2h^2}{8ma^2} \).

X-Rays

X-rays are produced when fast electrons strike a heavy metal target of high melting point. The continuous spectrum has a sharp short-wavelength limit \( \lambda_{min}=\frac{hc}{eV}=\frac{12400}{V} \) Å, depending only on the tube voltage. The tube current controls the intensity (the number of electrons striking per second is I/e). Superimposed on it are the characteristic lines, which arise when an inner-shell vacancy is filled; they identify the element and need the incident electron energy to exceed the inner-shell binding energy.
The line names follow the shell reached: \( K_\alpha \) (2 to 1), \( K_\beta \) (3 to 1), \( L_\alpha \) (3 to 2). Energies add, so \( f_{K_\beta}=f_{K_\alpha}+f_{L_\alpha} \). Moseley's law for the K-alpha line is \( \sqrt f=a(Z-1) \), so \( \lambda\propto\frac{1}{(Z-1)^2} \).

Worked Examples

Light of 300 nm falls on a metal of work function 2.5 eV: the photon energy is 1240/300 = 4.13 eV, so \( K_{max}=1.63 \) eV and \( V_0=1.63 \) V.
An electron accelerated through 150 V has \( \lambda\approx\frac{12.27}{\sqrt{150}}=1.0 \) Å.
An X-ray tube at 31 kV has \( \lambda_{min}=\frac{12400}{31000}=0.4 \) Å.
The K-alpha wavelengths of elements with Z = 11 and Z = 6 are in the ratio \( \frac{\lambda_6}{\lambda_{11}}=\frac{(11-1)^2}{(6-1)^2}=4 \).