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A photon is incident having frequency $1 \times 10^{14} \mathrm{sec}^{-1}$. Threshold frequency of metal is $5 \times 10^{13} \mathrm{sec}^{-1}$. Find the kinetic energy of the ejected electron.
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Verified Answer
The correct answer is:
$3.3 \times 10^{-20} \mathrm{~J}$
The relation between kinetic energy,
threshold frequency and frequency of photon is given as
$\frac{1}{2} m v^2=\left(h v-h v_0\right)$
$=h\left(1 \times 10^{14}-5 \times 10^{13}\right)$
$k=h\left(5 \times 10^{13}\right)$
$=6.67 \times 10^{-34} \times 5 \times 10^{13}$
$=3.3 \times 10^{-20} \mathrm{~J}$
threshold frequency and frequency of photon is given as
$\frac{1}{2} m v^2=\left(h v-h v_0\right)$
$=h\left(1 \times 10^{14}-5 \times 10^{13}\right)$
$k=h\left(5 \times 10^{13}\right)$
$=6.67 \times 10^{-34} \times 5 \times 10^{13}$
$=3.3 \times 10^{-20} \mathrm{~J}$
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