Calculate the energy emitted when electrons of $2.0 g$ atom of hydrogen undergo transition giving the spectral lines of lowest energy in the visible region of its atomic spectra.
$R_{H} = 1.1 \times 10^{7} m^{- 1}; C = 3 \times 10^{8} m / \sec; h = 6.625 \times 10^{- 34} J . \sec$

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$182.5 K J$

$365 K J$

$91.25 K J$

$18.25 K J$

answer is B.

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Detailed Solution

$\frac{1}{λ} = R_{H} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] n_{1} = 2 n_{2} = 3 λ = 6.55 \times 10^{- 7} m E = \frac{hc}{λ} . \frac{6.625 \times 10^{- 34} \times 3 \times 10^{8}}{6.55 \times 10^{- 7}} = 3.03 \times 10^{- 19} 5 Energy released = E \times N = 3.03 \times 10^{- 19} \times 6.023 \times 10^{23} \times 2 = 365 KJ$