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?

Answer 1

let's find energy of each photon of given ultraviolet light.

E = hc/λ

here, h = 6.63 × 10^-34J.s, c = 3 × 10^8 k/s and λ = 2271A° = 2271 × 10^-10 m

(6.63 × 10^-34 × 3 × 10^8)/(2271 × 10^-10 × 1.6 × 10^-19) eV [ as you know, 1eV = 1.6 × 10^-19 J ]

= 5.47eV

now, maximum kinetic energy of emitted electron = charge on electron × stopping potential

= 1e × 1.3V [ given stopping potential = 1.3 V]

or, $\frac{1}{2}mv_{max}^2=1.3eV$

using photoelectric equation,

$E=W_0+\frac{1}{2}mv_{max}^2$

or, 5.47eV = $W_0$ + 1.3eV

or, $W_0$ = 5.47eV - 1.3eV = 4.17eV

red light of wavelength 6328A° will have energy of each photon, E = hc/λ

= (6.63 × 10^-34 × 3 × 10^8)/(6328 × 10^-10 × 1.6 × 10^-19) = 1.96eV

Thus, energy of red light photons is less than work function (4.17 > 1.96) hence, irrespective of any intensity, no emission will take place.