Question

calculate wavelength of radiation emitted when electron de exists from n=2 to n=1 hydrogen atom

jisko acha sa ata ha wo hi karna kuki ha board ka ha islaya​

Answer 1

Answer:

see the attached pic

Explanation:

i hope it's helpful

Answer 2

Answer:

Given,

$n_1$ = 1

$n_2$ =2

we know that,

$R_H$ = 109678 (cm)^-1

wavelength $\lambda$ = to find

For hydrogen atom,

$\sf \: \frac{1}{ \lambda} = R_H \times ( \frac{1}{ {n_1}^{2} } - \frac{1}{{n_2}^{2}} ) \\ \sf \: \: \: \: \: \: = 109678 \times (\frac{1}{ {1}^{2} } - \frac{1}{ {2}^{2} } ) \\ \sf \: \: \: \: \: \: = 109678 \times ( \frac{1 }{1} - \frac{1}{4} ) \\ \sf \: \: \: \: \: \: = 109678 \times \frac{3}{4} \\ \sf \implies \frac{1}{ \lambda} = 82258.5 \\ \sf \implies \lambda \: = \frac{1}{82258.5} \\ \sf \: \: \: \: \: \: \: \: \: \: = 1.215 \times {10}^{ - 5} \: \: cm \\ \sf \: \: \: \: \: \: \: \: \: = 1.215 \times {10}^{ - 5} \times {10}^{7} \: \: nm \: \\ \sf \: \: \: \: \: \: \: \: \: \: = 121.5 \: \: nm$