Question

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

Answer 1

The correct option is B $1.41\text{eV}$

Initially, energy of electron $=+3\text{eV}$
Finally, in $2^{nd}$ excited state, energy of electron is given as
$-\frac{13.6\text{eV}}{n^2}=-\frac{\left(13.6\text{eV}\right)}{3^2}=-1.51\text{eV}..\left(1\right)$ [As $n=3$ for $2^{nd}$ excited state]
So, the energy of resulting photon will be
$\frac{hc}{\lambda }=(3-(-1.51\left)\right)\text{eV}=4.51\text{eV}..\left(2\right)$

Now, from photoelectric effect equation we have
$K.E_{max}=\frac{hc}{\lambda }-\varphi =4.51-\left(\frac{hc}{{\lambda }_{\text{thresold}}}\right)$
$=4.51-\frac{12400}{4000}=1.41\text{eV}$