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Q.

In an experiment of photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.12 \times 10^{- 15}$ V-s. Calculate the value of Planck’s constant

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a

$6 \times 10^{- 34} J s$

b

$6.99 \times 10^{- 34} J s$

c

$6.59 \times 10^{- 34} J s$

d

$7.2 \times 19^{- 34} J s$

answer is C.

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Detailed Solution

Slope of graph between cut-off voltage and frequency is give a by, slope = $\frac{h}{e}$ = $4.12 \times 10^{- 15} = \frac{h}{1.6 \times 10^{- 19}}$ ;

$h = 4.12 \times 1.6 \times 10^{- 34}$ = $6.59 \times 10^{- 34}$ Js

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In an experiment of photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12×10-15 V-s. Calculate the value of Planck’s constant