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  • The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength <img alt="\mathrm{6630 \, \AA \, is \: 0.42 \mathrm{~V}}" src="https://entrancecorner.oncodecogs.

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength \mathrm{6630 \, \AA \, is \: 0.42 \mathrm{~V}}. If the threshold frequency is \mathrm{x \times 10^{13} / \mathrm{s}} , where \mathrm{x} is______ (nearest integer).
 (Given, speed light \mathrm{=3 \times 10^{8} \mathrm{~m} / \mathrm{s}}, Planck's constant \mathrm{=6.63 \times 10^{-34} \mathrm{Js}} )

Option: 1

35


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{\lambda=6630 \AA}
\mathrm{V_{0}=0.42 \mathrm{~V}}
\mathrm{f_{0}=\underset{\text { }}{\text { threshold frequency }}=x \times 10^{13}}
\mathrm{Energy \, of\: incident \: photon =\frac{12400 *}{6630}(\mathrm{eV})}
                                                      \mathrm{h f=1.87 \mathrm{eV}}
 

By Einstein's photodectric equation,
\mathrm{h f =\phi_{0}+e V_{0}}
\mathrm{\phi_{0} =h f_{0}=1.87 \mathrm{eV}-0.42 \mathrm{eV}}
               \mathrm{f_{0} =\frac{1.45 \mathrm{eV}}{\mathrm{h}}}

\mathrm{f_{0} =\frac{1.45 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}}}
      \mathrm{=0.35 \times 10^{15}}
\mathrm{f_{0} =35 \times 10^{13} \mathrm{~Hz}}

\mathrm{\therefore x =35}

Posted by

seema garhwal

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