In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to : \left(\right. \frac{hc}{e} = 1240 \textrm{ }\textrm{ } nm

Engineering

Physics

Photoelectric Effect Einsteins Equation

Question

In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to : $(\frac{hc}{e} = 1240 nm - V)$

0.5 V

1.5 V

2.0 V

1.0 V

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$\frac{hc}{λ_{1}} = ϕ + {eV}_{1} . ..... (i)$

$\frac{hc}{λ_{2}} = ϕ + {eV}_{2} . ..... (ii)$

(i) – (ii)

$hc (\frac{1}{λ_{1}} - \frac{1}{λ_{2}}) = e (v_{1} - V_{2})$

$\Rightarrow V_{1} - V_{2} = \frac{hc}{e} (\frac{λ_{2} - λ_{1}}{λ_{1} - λ_{2}})$

$= (1240 nm V) \frac{100 nm}{300 nm \times 400 nm}$

$= \frac{12 .4}{12} \approx 1 V .$