Question: Q. 3. (i) Using Bohr’s postulates, derive the expression for the total energy of the electron in the stationary states of the hydrogen atom.
U] [Foreign, OD Comptt. I, II, III 2014;
Delhi I, II, III 2013]
(ii) Using Rydberg formula, calculate the wavelengths of the spectral lines of the first member of the Lyman series and of the Balmer series.
[Foreign 2014]
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Solution:
Ans. (i) } \quad \begin{aligned} m v r & =\frac{n h}{2 \pi} \ \frac{m v^{2}}{r} & =\frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r^{2}} \ r & =\frac{e^{2}}{4 \pi \varepsilon_{0} m v^{2}} \ r & =\frac{\mathrm{Z} e^{2}}{4 \pi \varepsilon_{0} m\left(\frac{n h}{2 \pi m r}\right)^{2}} \ \Rightarrow \quad r & =\frac{\varepsilon_{0} n^{2} h^{2}}{\pi m e^{2}} \end{aligned} $$
$$ \begin{aligned} & \text { Potential energy, } U=-\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{e^{2}}{r} \ & =-\frac{m e^{4}}{4 \varepsilon_{0} n^{2} h^{2}} \ & \text { K.E. }=\frac{1}{2} m v^{2}=\frac{1}{2} m\left(\frac{n h}{2 \pi m r}\right)^{2} \quad 1 / 2 \ & =\frac{n^{2} h^{2} \pi^{2} m^{2} e^{4}}{8 \pi^{2} m e_{0}^{2} n^{4} h^{4}} \ & \text { K.E. }=\frac{m e^{4}}{8 \varepsilon_{0}^{2} n^{2} h^{2}} \ & \text { T.E. = K.E. + P.E. } \ & =-\frac{m e^{4}}{8 \varepsilon_{0}^{2} n^{2} h^{2}} \end{aligned} $$
(ii) Rydberg formula : For first member of Lyman series
$$ \begin{aligned} \frac{1}{\lambda} & =R\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right) \ \lambda & =\frac{4}{3 R}=\frac{4}{3} \times 912 \AA \ & =1216 \AA \end{aligned} $$
For first member of Balmer Series
$$ \begin{aligned} \frac{1}{\lambda} & =R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right) \ \lambda & =\frac{36}{5 R} \ & =\frac{36}{5} \times 912 \AA \ & =6566 \cdot 4 \AA \end{aligned} $$
[CBSE Marking Scheme 2014]
