JEE Main · ChemistryMedium

$Calculate the wavelength of the photon emitted when an electron in a He^{+} ion transitions from the n = 3 to the n = 2 energy level. (Given: Rydberg constant R = 1.097 \times 1 0^{7} m^{- 1})$

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Correct answer: A — $1.64 \times 1 0^{- 7} m$

Solution

$The Rydberg formula for a hydrogen-like ion is \frac{1}{λ} = R Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}) .$
$For He^{+}, Z = 2, n_{1} = 2, and n_{2} = 3 .$
$\frac{1}{λ} = 1.097 \times 1 0^{7} \times 2^{2} \times (\frac{1}{4} - \frac{1}{9}) = 1.097 \times 1 0^{7} \times 4 \times \frac{5}{36} .$
$\frac{1}{λ} = 1.097 \times 1 0^{7} \times \frac{5}{9} \approx 6.094 \times 1 0^{6} m^{- 1} .$
$λ \approx 1.64 \times 1 0^{- 7} m . Hence the answer is (A).$