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The stopping potential \mathrm{V} for photoelectric emission from a metal surface is plotted along Y-axis and frequency \mathrm{\nu} of incident light along X-axis. A straight line is obtained as shown in the figure. Planck’s constant is given by:

Option: 1

slope of the line
 


Option: 2

 product of slope of the line and charge on the electron
 


Option: 3

ntercept along Y-axis divided by charge on the electron
 


Option: 4

product of intercept along X-axis and mass of the electron


Answers (1)

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According to Einstein's photoelectric equation
\mathrm{K}=\mathrm{eV}=\mathrm{h\nu}-\mathrm{W}_0 \quad \text { or } \quad \mathrm{V}=\frac{\mathrm{h}}{\mathrm{e}} \mathrm{\nu}-\frac{\mathrm{W}_0}{\mathrm{e}}
\therefore  Slope of straight line between \mathrm{V} and \mathrm{v} \text{ is } \frac{\mathrm{h}}{\mathrm{e}}.

\therefore \mathrm{h}=\mathrm{e} \times slope of the straight line.

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Ritika Harsh

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