Question

In a photoelectric experiment, the emitter plate is irradiated with radiation of 200 nm. The photocurrent becomes zero when the collector plate potential is −0.80 V. Calculate the work function (in eV) of the emitter.

Answer 1

Given: Wavelength (λ) = 200 nm

Stopping potential V_s = 0.80 V

The energy of the incident radiation (E) is determined by its wavelength (λ). Using the relationship between energy, Planck’s constant (k), and the speed of light (c):

E = `(h c)/lambda`

Using the common approximation (hc) = 1240 eV.nm:

E = `1240/200`

= 6.20 eV

The photocurrent becomes zero at the stopping potential (V₀ = 0.80 V). This potential corresponds to the maximum kinetic energy (K_max) of the emitted photoelectrons:

K_max = e . V₀

= 0.80 eV

According to Einstein’s equation, the energy of the incident photon is the sum of the work function (Φ) and the maximum kinetic energy of the electrons:

E = Φ + K_max

Φ = E − K_max

= 6.20 − 0.80

= 5.40 eV

∴ The work function of the emitter is 5.40 eV.