Question 565: Atoms & Nuclei

Question

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

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