Monochromatic radiation of wavelength 640.2 nm (1nm = 10−9 m) from a neon lamp irradiates photosensitive material made of caesium on tungsten.

Solution:

Wavelength of the monochromatic radiation, λ = 640.2 nm

= 640.2 × 10−9 m

Stopping potential of the neon lamp, V0 = 0.54 V

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

Planck’s constant, h = 6.6 × 10−34 Js

Let $\phi_{0}$ be the work function and $v$ be the frequency of emitted light.

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

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

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

$=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{640.2 \times 10^{-9}}-1.6 \times 10^{-19} \times 0.54$

$=3.093 \times 10^{-19}-0.864 \times 10^{-19}$

$=2.229 \times 10^{-19} \mathrm{~J}$

=\frac{2.229 \times 10^{-19}}{1.6 \times 10^{-19}}=1.39 \mathrm{eV}

Wavelength of the radiation emitted from an iron source, λ' = 427.2 nm

= 427.2 × 10−9 m

Let $V_{0}^{\prime}$ be the new stopping potential. Hence, photo-energy is given as:

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

$=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{427.2 \times 10^{-9}}-2.229 \times 10^{-19}$

$=4.63 \times 10^{-19}-2.229 \times 10^{-19}$

$=2.401 \times 10^{-19} \mathrm{~J}$

$=\frac{2.401 \times 10^{-19}}{1.6 \times 10^{-19}}=1.5 \mathrm{eV}$

Hence, the new stopping potential is 1.50 eV.