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If an electron in a hydrogen atom jumps from the 3^rd orbit to the 2^nd orbit, it emits a photon of wavelength $λ$ . When it jumps from the 4^th orbit to the 3^rd orbit, the corresponding wavelength of the photon will be:

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$\frac{20}{13} λ$

$\frac{16}{25} λ$

$\frac{9}{16} λ$

$\frac{20}{7} λ$

answer is D.

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Detailed Solution

From Rydberg’s Equation

$E = h \times C \times R_{H} (\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{p}}^{2}})$

$R_{H} = 1.09 \times 10^{7} m^{- 1}$

$n f = 3 n i = 1$

$E = 6.6 \times 10^{- 34} \times 3.0 \times 10^{8} \times 1.09$

$E = 19.30 \times 10^{- 19} J$

$E = h \times v$

$19.30 \times 10^{- 19} J = 6.6 \times 10^{- 34} J s \times v$

$v = 2.92 \times 10^{15} s^{- 1}$

$λ^{1} = \frac{1}{λ [\frac{1}{{(4)}^{2}} - \frac{1}{{(3)}^{2}}]}$

$λ^{1} = λ [\frac{1}{\frac{7}{20}}]$

$λ^{1} = \frac{20}{7} λ$

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