QJEEM19Ch3af3: JEE Main 2019 Chemistry - Class 11 JEEM19Ch3af3 Solution: If p is the momentum of the fastest electron ejected from a…

From photoelectric effect,

E = $\phi$ + KE

${{hc} \over \lambda } = \phi + {{{p^2}} \over {2m}}$ ................ (1)

Now when momentum = 1.5p then let wavelength = ${{\lambda _1}}$

$\therefore$ ${{hc} \over {{\lambda _1}}} = \phi + {{{{\left( {1.5p} \right)}^2}} \over {2m}}$ ........... (2)

Given,

kinetic energy(KE) of ejected photoelectron to be very high in comparison to work function( $\phi$ ).

$\therefore$ We can neglect work function( $\phi$ ).

$\therefore$ Equation (1) and (2) becomes,

${{hc} \over \lambda } = {{{p^2}} \over {2m}}$ ................ (1)

${{hc} \over {{\lambda _1}}} = {{{{\left( {1.5p} \right)}^2}} \over {2m}}$ ........... (2)

Dividing (1) by (2) we get,

${{{\lambda _1}} \over \lambda } = {{{p^2}} \over {{{\left( {1.5p} \right)}^2}}}$

$\Rightarrow$ ${{{\lambda _1}} \over \lambda } = {4 \over 9}$

$\Rightarrow$ ${\lambda _1} = {4 \over 9}\lambda$