Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal.
Question:

Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is −1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (105 W m−2) red light of wavelength 6328 Å produced by a He-Ne laser?

Solution:

Wavelength of ultraviolet light, λ = 2271 Å = 2271 × 10−10 m

Stopping potential of the metal, V0 = 1.3 V

Planck’s constant, h = 6.6 × 10−34 J

Charge on an electron, e = 1.6 × 10−19 C

Work function of the metal $=\phi_{0}$

Frequency of light = ν

We have the photo-energy relation from the photoelectric effect as

$\phi_{0}=h v-e V_{0}$

$=\frac{h c}{\lambda}-e V_{0}$

$=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{2271 \times 10^{-10}}-1.6 \times 10^{-19} \times 1.3$

$=8.72 \times 10^{-19}-2.08 \times 10^{-19}$’

$=6.64 \times 10^{-19} \mathrm{~J}$

$=\frac{6.64 \times 10^{-19}}{1.6 \times 10^{-19}}=4.15 \mathrm{eV}$

Let ν0 be the threshold frequency of the metal.

$\therefore \phi_{0}=h v_{0}$

$v_{0}=\frac{\phi_{0}}{h}$

$=\frac{6.64 \times 10^{-19}}{6.6 \times 10^{-34}}=1.006 \times 10^{15} \mathrm{~Hz}$

Wavelength of red light, $\lambda_{r}=6328 \AA=6328 \times 10^{-10} \mathrm{~m}$

$\therefore$ Frequency of red light, $v_{r}=\frac{c}{\lambda_{r}}$

$=\frac{3 \times 10^{8}}{6328 \times 10^{-10}}=4.74 \times 10^{14} \mathrm{~Hz}$

Since ν0νr, the photocell will not respond to the red light produced by the laser.