JEE Main · ChemistryHard

$What is the frequency ν of the radiation emitted when an electron in a hydrogen atom jumps from the n = 3 to the n = 2 energy level? (Given R_{H} = 1.097 \times 1 0^{7} m^{- 1} and c = 3 \times 1 0^{8} m/s)$

Show correct answer & step-by-step solution

Correct answer: A — $4.57 \times 1 0^{14} Hz$

Solution

$The wavenumber is \overset{ν}{ˉ} = R_{H} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}) = 1.097 \times 1 0^{7} (\frac{1}{2 ^{2}} - \frac{1}{3 ^{2}}) m^{- 1} .$
$\overset{ν}{ˉ} = 1.097 \times 1 0^{7} (\frac{1}{4} - \frac{1}{9}) = 1.097 \times 1 0^{7} (\frac{5}{36}) \approx 1.5236 \times 1 0^{6} m^{- 1} .$
$Frequency ν = c \cdot \overset{ν}{ˉ} = 3 \times 1 0^{8} \cdot 1.5236 \times 1 0^{6} \approx 4.57 \times 1 0^{14} Hz .$
$Hence the answer is (A).$