# 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)$\frac{4}{9}\lambda$

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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