Question

In a photoelectric-effect experiment it is observed that no current flows unless the wavelength is less than 570 nm .
A) What is the work function of this material?
B) What is the stopping voltage required if light of wavelength 420 nm is used?

Answer 1

Answer:

$2.17960625\ eV$

0.77842 V

Explanation:

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

c = Speed of light = $3\times 10^8\ m/s$

$\lambda$ = Wavelength

Work function is given by

$W=\dfrac{hc}{\lambda}\\\Rightarrow W=\dfrac{6.626\times 10^{-34}\times 3\times 10^8}{570\times 10^{-9}}\\\Rightarrow W=3.48737\times 10^{-19}\ J=\dfrac{3.48737\times 10^{-19}}{1.6\times 10^{-19}}=2.17960625\ eV$

The work function is $2.17960625\ eV$

Stopping voltage is given by

$eV_0=hf-W_0\\\Rightarrow eV_0=\dfrac{6.626\times 10^{-34}\times 3\times 10^8}{1.6\times 10^{-19}\times 420\times 10^{-9}}-2.17960625\\\Rightarrow eV_0=0.77842\ eV\\\Rightarrow V_0=0.77842\ V$

The stopping voltage required is 0.77842 V

Answer 2

Answer:

(a) 3.49 x 10^-19 J

(b) 0.78 V

Explanation:

threshold wavelength, λo = 570 nm = 570 x 10^-9 m

(A) the work function of the metal is given by

$\phi =\frac{hc}{\lambda _{0}}$

where h is the Plank's constant and c be the speed of light.

h = 6.63 x 10^-34 Js

c = 3 x 10^8 m/s

$\phi =\frac{6.63\times 10^{-34}\times 3 \times 10^{8}}{570\times 10^{-9}}$

Ф = 3.49 x 10^-19 J

(B) λ = 420 nm = 420 x 10^-9 m

Use Einstein relation

$\frac{hc}{\lambda }=\phi +eV_{0}$

where, Vo is the stopping potential

$\frac{6.63\times10^{-19}\times 3\times 10^{8}}{420\times 10^{-9} }=3.49 \times 10^{-19} +eV_{0}$

eVo = 1.246 x 10^-19

Vo = 0.78 V