Question

An electron and proton are separated by a large distance. The electron starts approaching the proton with energy $3 \,eV$. 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. $7.61\, eV$
• B. $1.41 \,eV$
• C. $3.3 \,eV$
• D. No photoelectron would be emitted

Answer 1

Answer: (B)

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