Question 521: Atoms & Nuclei

Question

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

Accepted Answer

When electron jumps from higher orbit to
lower orbit then, wavelength of emitted photon is
given by,
$${1 \over \lambda } = R\left( {{1 \over {n_f^2}} - {1 \over {n_i^2}}} ight)$$
On jumping from 3
rd
orbit to 2
nd
orbit,
$${1 \over \lambda } = R\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} ight)$$ = $${{5R} \over {36}}$$
On jumping from 4
th
orbit to 3
rd
orbit,
$${1 \over {\lambda '}} = R\left( {{1 \over {{3^2}}} - {1 \over {{4^2}}}} ight)$$ = $${{7R} \over {144}}$$
$$ 	herefore $$ $$\lambda $$' = $${{144} \over 7} 	imes {{5\lambda } \over {36}}$$ = $${{20\lambda } \over 7}$$

Answer

$${{16} \over {25}}\lambda $$

Answer

$${9 \over {16}}\lambda $$

Answer

$${{20} \over 7}\lambda $$

Answer

$${{20} \over {13}}\lambda $$

Atoms & Nuclei

If an electron in a hydrogen atom jumps from the 3 rd orbit to the 2 nd orbit, it emits a photon of wavelength $\lambda $. 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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