Question

The work function of a photoelectric material is 4.0 eV. (a) What is the threshold wavelength? (b) Find the wavelength of light for which the stopping potential is 2.5 V.

Answer 1

(a) The threshold wavelength is $310 \mathrm{nm}$

(b) The wavelength of light for which the stopping potential is 2.5 V is 191 nm .

Explanation:

A photoelectric material works function ϕ = 4 eV = $4 \times 1.6 \times 10^{-19} \mathrm{J}$

Potential ends, $V_0$ = 2.5 V

Planck's constant, h = 6.63 × 10−34 Js  

(a) To find the threshold wavelength:

A photoelectric material Work function ,

$\phi=\frac{h c}{\lambda_{0}}$  

Here, λ0 = Light wavelength threshold

            c = velocity of light

$\lambda_{0}=\frac{h c}{\phi}$

    $=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\left(4 \times 1.6 \times 10^{-19}\right)}$

$\lambda_{0}=\frac{6.63 \times 3}{64} \times \frac{10^{-27}}{10^{-9}}$

$\lambda_{0}=3.1 \times 10^{-7} \mathrm{m}$

$\lambda_{0}=310 \mathrm{nm}$

(b) To find the wavelength of light for which the stopping potential is 2.5 V:

From the Photoelectric Equation of Einstein,

$E=\phi+e V_{0}$

When the respective values are replaced, we get:

$\frac{h c}{\lambda}=4 \times 1.6 \times 10^{-19}+1.6 \times 10^{-19} \times 2.5$

$\lambda=\frac{\left(6.63 \times 10^{-34} \times 3 \times 10^{8}\right)}{\left(6.5 \times 1.6 \times 10^{-19}\right)}$

$\lambda=\frac{6.63 \times 3 \times 10^{-26}}{1.6 \times 10^{-19} \times 6.5}$

$\lambda=\frac{6.63 \times 3 \times 10^{-26}}{1.6 \times 10^{-19} \times 6.5}$

$\lambda=1.9125 \times 10^{-7}$

  = 191 nm