Question

A difference of $2.3 \, eV$ separates two energy levels in an atom. The frequency of radiation emitted when the electron makes a transition is

• A. $6.7\times 10^{14}Hz$
• B. $4.6\times 10^{14}Hz$
• C. $5.6 \, \times \, 10^{14} \, Hz$
• D. $7.7\times 10^{14}Hz$

Answer 1

Answer: (C)

Given, the difference in energy level $E \, = \, 2.3 \, eV$
$ \, = \, 2.3 \, \times \, 1.6 \, \times \, 10^{- 19} \, J$
​Planck's constant $h \, = \, 6.63 \, \times \, 10^{- 34} \, J \, s$
Let $v$ be the frequency, then $E \, = \, hf$
or $f=\frac{E}{h}=\frac{\text{2.3} \times \text{1.6} \times \text{10}^{- \text{19}}}{\text{6.63} \times \text{10}^{- \text{34}}}$
$= \, 5.6 \, \times \, 10^{14} \, Hz \, \, $