Find the order of magnitude of frequency of light which ejects electrons from a metal surface. Fully stopped by a retarding potential of 3V, the photoelectric effect begins in this metal at a frequency of $6 \times 10^{14} H z$ . (Given $h = 6.63 \times 10^{- 34} J s$ ).

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Solution

According to Einstein's photoelectric equation, $E_{k} = h v - W$
If $V_{s}$ is the retarding or stopping potential and $v_{0}$ the threshold frequency, then above equation becomes
$e V_{x} = h v - h v_{0}$
or $h v = e V_{s} +_{h} v_{0}$
or $v = \frac{e V_{s}}{h} + v_{0}$
Hence, $e = 1.6 \times 10^{- 10} c o u l o m b, V_{s} = 3 V$ , and $v_{0} = 6 \times 10^{14} H z$
Required frequency
$v = \frac{1.6 \times 10^{- 10} \times 3}{6.63 \times 10^{- 34}} + 6 \times 10^{14}$
$= 7.24 \times 10^{14} + 6 \times 10^{14}$
$= 13.24 \times 10^{14} H z$
$= 1.324 \times 10^{15} H z$

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