The figure shows the plot of the stopping potential versus the frequency of the light used in an experiment on photoelectric effect. The ratio h/e is

10^{- 15} V s^{- 15} V s

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2 \times 10^{- 15} V s

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3 \times 10^{- 15} V s

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4.14 \times 10^{- 15} V s

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Solution

The correct option is D $4.14 \times 10^{- 15} V s$
The Einstein's equation for photoelectric effect is given by $K = h f - w$
or $e V_{s} = h f - w$ where $h =$ Planck's constant , $w =$ work function for metal , $f =$ frequency of incident photon.
So, $V_{s} = (h / e) f - w / e$
It will represent a straight line with slope $m = h / e$ .

From the given plot, $m = h / e = \frac{1.656}{4 \times 10^{14}} = 4.14 \times 10^{- 15} V - s$

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