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  • The kinetic energy of emitted electron is when the light incident on the metal has wavelength <img al

The kinetic energy of emitted electron is \mathrm{E} when the light incident on the metal has wavelength \lambda. To double the kinetic energy, the incident light must have wavelength :
 

Option: 1

\mathrm{\frac{h c}{E \lambda-h c}}
 


Option: 2

\frac{\text { hc } \lambda}{\mathrm{E} \lambda+\mathrm{hc}}


Option: 3

\mathrm{\frac{h \lambda}{\mathrm{E} \lambda+\mathrm{hc}}}


Option: 4

\mathrm{\frac{h c \lambda}{\text { E } \lambda-h c}}


Answers (1)

best_answer

From Einstein's Photoelectric effect,

\mathrm{\frac{hc}{\lambda }=\phi _{o}+E}\rightarrow (1)

Let the wavelength be \mathrm{\lambda '},when the kinetic energy is \mathrm{2E}

\mathrm{\frac{h c}{\lambda^{\prime}}=\phi_0+2 E \rightarrow 2 }

Eq. (2) - (1)

\mathrm{\frac{h c}{\lambda^{\prime}} -\frac{h c}{\lambda}=E }

\mathrm{\frac{h c}{\lambda^{\prime}} =E+\frac{h c}{\lambda} }

\mathrm{\lambda^{\prime} =\frac{h c \lambda}{E \lambda+h c}}

Posted by

sudhir.kumar

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