If P is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength \lambda,  then for 1.5 p momentum of the photoelectron, the wavelength of the light should be:

(Assume kinetic energy of elected photoelectron to be very high in comparison to work function):

  • Option 1)

    \frac{3}{4}\lambda

  • Option 2)

    \frac{1}{2}\lambda

  • Option 3)

    \frac{2}{3}\lambda

  • Option 4)

    \frac{4}{9}\lambda

Answers (1)

given    \frac{1}{2}mv^{2}>hv_{0}

               (K.E)            (work function)

\therefore hv=\frac{1}{2}mv^{2}

\frac{hc}{\lambda}=\frac{1}{2}mv^{2}=\frac{1}{2}mc^{2}

\lambda=2\frac{h}{mv}=\frac{2h}{p}-(i)

\lambda'=\frac{2h}{1.5p}-(ii)

\frac{\lambda}{\lambda'}=\frac{\frac{2h}{p}}{\frac{2h}{1.5p}}=1.5\Rightarrow \lambda'=\frac{2}{3}\lambda

 


Option 1)

\frac{3}{4}\lambda

Option 2)

\frac{1}{2}\lambda

Option 3)

\frac{2}{3}\lambda

Option 4)

\frac{4}{9}\lambda

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