Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (ν0 ) and work function (W0 ) of the metal.

Asked by Pragya Singh | 1 year ago |  111

1 Answer

Solution :-

Threshold wavelength of the radiation 

\((\lambda _{0}= 6800 Å\)\(\)

\( 6800\times 10^{-10}\, m)\)

Threshold frequency of the metal 

\(\nu _{0} =\dfrac{c}{\lambda _{0}}\)

\( =\dfrac{3\times 10^{8}ms^{-1}}{6.8\times 10^{-7}m}\)

\( =4.41\times 10^{14}\, s^{-1}\)

Therefore, threshold frequency \(\nu _{0}\) of the metal 

 \(=h\nu _{0}\\ \\ \)

\( =(6.626\times 10^{-34}Js)(4.41\times 10^{14}s^{-1})\\ \\ \)

\( =2.922\times 10^{-19}\, J\)

Answered by Pragya Singh | 1 year ago

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