The work function for the following metals is given:
Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV.
Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 A from a He-Cd laser placed 1m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away?

Solution

Given,
Work function of Na is,

ϕ_{Na} = 1.92 eV = 1.92 \times 1.6 \times 10^{- 19} J

Wavelength of the radiation light emitted,

λ = 3300 Å = 3300 \times 10^{- 10} m

Using the formula of energy of a photon,

E = \frac{hc}{λ} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{3300 \times 10^{- 10}} J

E = \frac{6.6 \times 3 \times 10^{- 34 + 8 + 10 - 2}}{33 \times 10} J

= \frac{6 \times 10^{- 18 - 1}}{1.6 \times 10^{- 19}} eV

= \frac{60}{10} eV

E = 3.75 eV

It is observed that energy of incident radiation is less thanwork function of Ni and Mo but larger than the work function of Na and K.
So, photoemission current take place from Na and K but not from Mo and Ni.

The intensity of the radiation increases, if the laser is brought closer and the photoelectric current from Na and K will increase. But, however this will not bring any change in the result derived.

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