Photoelectric effect and study of photoelectric effect

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Question

Figure shows the stopping potential (V)versus frequency (v) graph for photoelectric effect in a metal. From the graph, The Planck's constant (h) is $y \times 10^{- 34} Js$ . The value of y is

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Solution

${eV}_{0} = h (ν - ν_{0}) = h ν - h ν_{0}$
$V_{0} = (\frac{h}{e}) ν - (\frac{h}{e}) v_{0}$ ………..(1)
From Eq. (1) it follows that the slope of the graph is $\frac{h}{e}$ . Therefore
$\begin{matrix} h = e \times slope \\ = (1 .6 \times 10^{- 19}) \times \frac{16 .5 - 0}{(8 - 4) \times 10^{15}} \\ = \frac{(1 .6 \times 10^{- 19}) \times 16 .5}{4 \times 10^{15}} \\ = 6 .60 \times 10^{- 34} Js \end{matrix}$

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Photoelectric effect and study of photoelectric effect

Remember concepts with our Masterclasses.

Figure shows the stopping potential (V)versus frequency (v) graph for photoelectric effect in a metal. From the graph, The Planck's constant (h) is y×10−34Js. The value of y is

eV0=hν−ν0=hν−hν0 V0=heν−hev0 ………..(1)From Eq. (1) it follows that the slope of the graph is he. Therefore h=e× slope =1.6×10−19×16.5−0(8−4)×1015=1.6×10−19×16.54×1015=6.60×10−34Js

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Figure shows the stopping potential (V)versus frequency (v) graph for photoelectric effect in a metal. From the graph, The Planck's constant (h) is $y \times 10^{- 34} Js$ . The value of y is