A difference of 2.21 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

$5.3 × 10^{14} Hz$

The correct answer is Option (1) → $5.3 × 10^{14} Hz$

Given:

Energy difference, $ΔE = 2.21\,eV$

$1\,eV = 1.6 \times 10^{-19}\,J$

Convert energy to joules:

$ΔE = 2.21 \times 1.6 \times 10^{-19} = 3.536 \times 10^{-19}\,J$

Formula:

$ν = \frac{ΔE}{h}$

where $h = 6.63 \times 10^{-34}\,J·s$

Substitute:

$ν = \frac{3.536 \times 10^{-19}}{6.63 \times 10^{-34}}$

$ν = 5.33 \times 10^{14}\,Hz$

Final Answer: Frequency = 5.33 × 10¹⁴ Hz