Question

Calculate the wave number and energy of the spectral line emitted by an excited hydrogen atom corresponding to n1=3 and n2=5.​

Answer 1

Answer:

1.55*10-¹⁹

Explanation:

E=hc/wavelength

6.6.26*10^-³⁴ *3*10⁸/12826*10¹⁰

=1.55*10^-19

Answer 2

Explanation:

$v' = R {z}^{2} ( \frac{1}{ {n1}^{2} } - \frac{1}{ {n2}^{2} } ) \\ v' = R \: {1}^{2}( \frac{1}{ {3}^{2} } - \frac{1}{ {5}^{2} } )$

R = 10967.8 / cm

$v' = 10967.8 ( \frac{25 - 5}{225} ) \\ v' = 10967.8 ( \frac{16}{225} ) \\ v' = 779.93 {cm}^{ - 1}$

we know that

$\frac{1}{λ} = v' \\ \frac{1}{λ} = 779.93 \\ λ = \frac{1}{779.93} \\ λ = 1.282 \times {10}^{ - 3} {cm}^{ - 1}$

using formula calculating energy

E = hc / λ

$E = \frac{hc}{ λ} \\ E = \frac{6.625 \times {10}^{ - 34} \times 3 \times {10}^{8} }{1.282 \times {10}^{ - 3} } \\E = \frac{19.875 \times {10}^{ - 42} }{1.282 \times {10}^{ - 3} } \\ E = 15.503 \times {10}^{39} J$

$so \\ wave \: number \: = 779.93 {cm}^{ - 1} \\ and \: energy \: = 15.503 \times {10}^{39} J$

disclaimer : major calculation are done by calculator for saving time .

hope it help