When a certain metal was irradiated with light of frequency $3.2 \times 10^{16} \mathrm{~Hz}$,…

When a certain metal was irradiated with light of frequency $3.2 \times 10^{16} \mathrm{~Hz}$, photoelectrons emitted had twice the kinetic energy as did photoelectrons emitted when the same metal was irradiated with light of frequency $2.0 \times 10^{16} \mathrm{~Hz}$. Hence, threshold frequency is
  1. $0.8 \times 10^{15} \mathrm{~Hz}$
  2. $8.0 \times 10^{15} \mathrm{~Hz}$
  3. $0.8 \times 10^{14} \mathrm{~Hz}$
  4. $6.4 \times 10^{16} \mathrm{~Hz}$

Solution

$\mathrm{hv}=\mathrm{hv}_{0}+(\mathrm{KE})$
$\mathrm{KE}=\mathrm{h}\left(\mathrm{v}-\mathrm{v}_{0}ight)$
$(\mathrm{KE})_{1}=\mathrm{h}\left(3.2 \times 10^{16}-\mathrm{v}_{0}ight)$
$(\mathrm{KE})_{2}=\mathrm{h}\left(2.0 \times 10^{16}-\mathrm{v}_{0}ight)$
Given $(\mathrm{KE})_{1}=2(\mathrm{~K} . \mathrm{E} .)_{2}$
$\mathrm{h}\left(3.2 \times 10^{16}-\mathrm{v}_{0}ight)=2 \mathrm{~h}\left(2.0 \times 10^{16}-\mathrm{v}_{0}ight)$
$\therefore \mathrm{v}_{0}=8.0 \times 10^{15} \mathrm{~Hz}$ ^

Asked in: JEE-TOPICTESTS-CHEMISTRY

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