find the wavelength of the radiation emitted by hydrogen in the following transitions (a)n=3 to n=2 (B)N=5 to n=4 and (c)n=10 to n=9 what is the order of the wavelengths obtained?

Dear Student,

Please find below the solution to the asked query.

$Wavenumber \left(\stackrel{-}{\nu }\right) = 1.09677\times {10}^7 (\frac{1}{n_i^2}-\frac{1}{n_f^2}) m^{-1}$
(i) n_i = 2
n_f = 3

So,
$$\begin{gathered} \stackrel{-}{\nu } = 1.09677\times {10}^7 (\frac{1}{2^2}-\frac{1}{3^2}) \\ = 1.09677\times {10}^7 (\frac{1}{4}-\frac{1}{9}) \\ = 1.09677\times {10}^7 \times \frac{5}{36} \\ = 1.52\times {10}^6 m^{-1} \end{gathered}$$

$Wavelength \left(\lambda \right) = \frac{1}{{\displaystyle \stackrel{-}{\nu }}}= \frac{1}{1.52\times {10}^6 m^{-1}} = 6.58\times {10}^{-7} m = 658 nm$

The order of this wavelength belongs to visible region.

(ii) n_i = 4
n_f = 5

So,
 $$\begin{gathered} \stackrel{-}{\nu } = 1.09677\times {10}^7 (\frac{1}{4^2}-\frac{1}{5^2}) \\ = 1.09677\times {10}^7 (\frac{1}{16}-\frac{1}{25}) \\ = 1.09677\times {10}^7 \times \frac{9}{25\times 16} \\ = 2.47\times {10}^5 m^{-1} \end{gathered}$$

$Wavelength \left(\lambda \right) = \frac{1}{{\displaystyle \stackrel{-}{\nu }}}= \frac{1}{2.47\times {10}^5 m^{-1}} = 4.05\times {10}^{-6} m$

The order of this wavelength comes under Infra red region.

(iii) n_i = 9
n_f = 10

So,

$$\begin{gathered} \stackrel{-}{\nu } = 1.09677\times {10}^7 (\frac{1}{9^2}-\frac{1}{{10}^2}) \\ = 1.09677\times {10}^7 (\frac{1}{81}-\frac{1}{100}) \\ = 1.09677\times {10}^7 \times \frac{19}{81\times 100} \\ = 2.57\times {10}^4 m^{-1} \end{gathered}$$

$Wavelength \left(\lambda \right) = \frac{1}{{\displaystyle \stackrel{-}{\nu }}}= \frac{1}{2.57\times {10}^4 m^{-1}} = 3.89\times {10}^{-5} m$

The order of this wavelength comes under infra red region.

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