JEE MAIN - Chemistry (2019 - 8th April Evening Slot - No. 3)
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 ejected
photoelectron to be very high in comparison
to work function)
1/2 $$\lambda $$
3/4 $$\lambda $$
4/9 $$\lambda $$
2/3 $$\lambda $$
Explanation
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,
$${{{\lambda _1}} \over \lambda } = {{{p^2}} \over {{{\left( {1.5p} \right)}^2}}}$$
$$ \Rightarrow $$ $${{{\lambda _1}} \over \lambda } = {4 \over 9}$$
$$ \Rightarrow $$ $${\lambda _1} = {4 \over 9}\lambda $$
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,
$${{{\lambda _1}} \over \lambda } = {{{p^2}} \over {{{\left( {1.5p} \right)}^2}}}$$
$$ \Rightarrow $$ $${{{\lambda _1}} \over \lambda } = {4 \over 9}$$
$$ \Rightarrow $$ $${\lambda _1} = {4 \over 9}\lambda $$
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