Question

An electron and proton are separated by a large distance. The electron starts approaching the proton with energy $3eV$. The proton captures the electrons and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength $4000\mathring{A}$. What is the maximum kinetic energy of the emitted photoelectron?

• A. $7.61eV$
• B. $1.41eV$
• C. $3.3eV$
• D. No photoelectron would be emitted

Answer 1

Answer: (B)

Initially, energy of electron $=+3eV$
finally, in $2^{\text{nd }}$ excited state,
energy of electron $=-\frac{(13.6eV)}{3^2}$
$=-1.51eV$
Loss in energy is emitted as photon,
So, photon energy $\frac{hc}{\lambda }=4.51eV$
Now, photoelectric effect equation
$K{E}_{\max }=\frac{hc}{\lambda }-\varphi =4.51-\left(\frac{hc}{{\lambda }_{th}}\right)$
$=4.51eV-\frac{12400eV\mathring{A}}{4000\mathring{A}}$
$=1.41eV$