Question

In an experiment on the photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.12 \, x \, 10^{- 15} \, V \, s$ . 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 $ $\frac{\text{h}}{\text{e}} = 4 \text{.} 1 2 \times 1 0^{- 1 5}$
h = 1.6 x 10^-19 x 4.12 x 10^-15
= 6.592 x 10^-34 J - s.