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Q. In case of photoelectric effect, slope of $V_{0}$ vs $\nu$ curve is (where $V_{0}$ - stopping potential, $\nu=$ subjected frequency)

NTA AbhyasNTA Abhyas 2022

Solution:

$h \nu=\omega_{0}+e V_{0} \ldots(1)$
$\omega_{0}=$ threshold energy
$h \nu=h \nu_{0}+e V_{0}$
$V_{0}=\frac{h \nu_{0}}{e}-\frac{h \nu_{0}}{e} \ldots(2)$
$\nu=$ subjected frequency
$\nu_{0}=$ threshold frequency
Comparing equation (2) with straight line equation:
$y = mx + c$
Slope of $\nu_{0} $ v/s $\nu$ curve $=\frac{ h }{ e }$