Question

A group of hydrogen atoms are prepared in n = 4 states. LKist the wavelength that are emitted as the atoms make transitions and return to n = 2 states.

Answer 1

Here $N_1=1,n_2=2$

$n_1=4\to 3\to 2$

$\Rightarrow \frac{1}{\lambda }=1.097\times {10}^7\left(\frac{1-4}{16}\right)$

$\Rightarrow \frac{1}{\lambda }=-\frac{1.097\times {10}^7\times 3}{16}$

$\Rightarrow \lambda =\frac{16\times {10}^{-7}}{3\times 1.097}$

= $-4.8617\times {10}^{-7}$

= $-486.1\times {10}^{-9}$

= - 487 nm

Again $n_1=4and{n}_2=3$

$\frac{1}{\lambda }=1.097\times {10}^7(\frac{1}{16}-\frac{1}{9})$

$\Rightarrow \frac{1}{\lambda }=1.097\times {10}^7\left(\frac{9-16}{144}\right)$

$\Rightarrow \frac{1}{\lambda }=\frac{1.097\times {10}^7\times 7}{144}$

$\Rightarrow \lambda =-\frac{144}{1.097\times {10}^7\times 7}$

= - 1875 nm

When transition is

$n_2=3to{n}_2=2$

$\frac{1}{\lambda }=1.097\times {10}^7(\frac{1}{9}-\frac{1}{4})$

$\Rightarrow \frac{1}{\lambda }=1.097\times {10}^7\left(\frac{4-9}{36}\right)$

$\Rightarrow \frac{1}{\lambda }=\frac{1.097\times {10}^7}{36}$

$\Rightarrow \lambda =\frac{36\times {10}^7}{1.097\times 5}=-656nm$