JEE MAIN - Physics (2021 - 27th July Evening Shift - No. 1)
An electron and proton are separated by a large distance. The electron starts approaching the proton with energy 3 eV. The proton captures the electron and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength 4000$$\mathop A\limits^o $$. What is the maximum kinetic energy of the emitted photoelectron?
7.61 eV
1.41 eV
3.3 eV
No photoelectron would be emitted
Explanation
initially, energy of electron = + 3eV
finally, in 2nd excited state,
energy of electron = $$ - {{(13.6eV)} \over {{3^2}}}$$
$$ = - 1.51eV$$
Loss in energy is emitted as photon,
So, photon energy $${{hc} \over \lambda } = 4.51eV$$
Now, photoelectric effect equation
$$K{E_{\max }} = {{hc} \over \lambda } - \phi = 4.51 - \left( {{{hc} \over {{\lambda _{th}}}}} \right)$$
$$ = 4.51eV - {{12400eV\mathop A\limits^o } \over {4000\mathop A\limits^o }}$$
$$ = 1.41eV$$
finally, in 2nd excited state,
energy of electron = $$ - {{(13.6eV)} \over {{3^2}}}$$
$$ = - 1.51eV$$
Loss in energy is emitted as photon,
So, photon energy $${{hc} \over \lambda } = 4.51eV$$
Now, photoelectric effect equation
$$K{E_{\max }} = {{hc} \over \lambda } - \phi = 4.51 - \left( {{{hc} \over {{\lambda _{th}}}}} \right)$$
$$ = 4.51eV - {{12400eV\mathop A\limits^o } \over {4000\mathop A\limits^o }}$$
$$ = 1.41eV$$
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