Question

10. Photocurrent recorded in the micro
ammeter in an experimental set-up of
photoelectric effect vanishes when the
retarding potential is more than 0.8 V if
the wavelength of incident radiation is
4950 Å. If the source of incident radiation
is changed, the stopping potential turns
out to be 1.2 V. Find the work function of
the cathode material and the wavelength
of the second source. solve this problem
​

Answer 1

Given:

V₀ = 0.8V, λ = 4950A⁰

V₀' = 1.2V, h = 6.63 × 10⁻³⁴Js, c = 3×10⁸ms⁻¹

To find:

Work function and wavelength

Solution:

firstly calculate the work function of cathode material,

so, formula used is

$\phi = \frac{hc}{\lambda } - V_0e$

$\phi = \frac{(6.63 \times 10^-^3^4)(3 \times 10^8)}{4.950 \times 10^-^7 } - (0.8)(1.6 \times 10^-^1^9)$

$\phi = 4.018 \times 10^-^1^9 - 1.28 \times 10^-^1^9$

$\phi = 2.738 \times 10^-^1^9 J$

now convert it into eV,

= $\frac{2.738 \times 10^-^1^9 J}{1.6 \times 10^-^1^9 J/eV}$

= 1.711eV

Wavelength can be calculated as,

$V_0'e = \frac{hc}{\lambda '}-\phi$

so, second source wavelength can be calculated as

$\lambda ' = \frac{hc}{V_0'e + \phi }$

$\lambda ' = \frac{(6.63 \times 10^-^3^4)(3 \times 10^8)}{(1.2)(1.6 \times 10^-^1^9) + 2.738 \times 10^-^1^9 }$

$\lambda ' = \frac{19.89 \times 10^2^6}{4.658 \times 10^-^1^9 }$

λ' = 4.270 × 10⁻⁷m

λ' = 4270A°

therefore, work function value is 1.711eV and value of wavelength is 4270A°.