Q. An electron passing through a potential difference of $4.9\ V$ collides with a memory atom and transfers it to the first excited state. What is the wavelength of a photon corresponding to the transition of the mercury atom to its normal state
A) $2050\ \mathring{\mathrm{A}}$
B) $2240\ \mathring{\mathrm{A}}$
C) $2525\ \mathring{\mathrm{A}}$
D) $2935\ \mathring{\mathrm{A}}$
Explanation
(c) $\frac{hc}{\lambda} = E = eV$
$\Rightarrow \lambda = \frac{hc}{eV} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19} \times 4.9} = 2525\ \mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.