Question

In an experiment on the photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.12x10^{-15}Vs$ . The value of Planck's constant is

• A. 6.592 x 10^-34 J s
• B. 6.592 x 10^-31 J s
• C. 9.592 x 10^-34 J s
• D. 6.592 x 10^-30 J s

Answer 1

Answer: (A)

Given, slope of graph tan $\theta$ = 4.12 x 10^-15 V - s and charge on electron e = 1.6 x 10^-19 C

For slope of graph $\text{tan }\theta =\frac{\text{V}}{\text{v}}$
We know that
hv = eV
$\frac{\text{V}}{\text{v}}=\frac{\text{h}}{\text{e}}$
$\therefore$ $\frac{\text{h}}{\text{e}}=4\text{.}12\times 10^{-15}$
h = 1.6 x 10^-19 x 4.12 x 10^-15
= 6.592 x 10^-34 J - s.