Question: Q. 10. Show that the radius of the orbit in hydrogen atom varies as $n^{2}$, where $n$ is the principal quantum number of the atom.
U] [Delhi I, II, III 2015]
$$ \begin{align*} \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{e^{2}}{r^{2}} & =\frac{m v^{2}}{r} \ r & =\frac{e^{2}}{4 \pi \varepsilon_{0} m v^{2}} \ m v^{2} r & =\frac{e^{2}}{4 \pi \varepsilon_{0}} \tag{i} \end{align*} $$
According to the Bohr’s postulate,
$$ \begin{align*} m v r & =\frac{n h}{2 \pi} \ m^{2} v^{2} r^{2} & =\frac{n^{2} h^{2}}{4 \pi^{2}} \end{align*} $$
Putting the value of $m v^{2} r$ from eqn. (i)
$$ \begin{aligned} \frac{e^{2}}{4 \pi \varepsilon_{0}} m r & =\frac{n^{2} h^{2}}{4 \pi^{2}} \ r & =\left(\frac{n^{2}}{m}\right)\left(\frac{h}{2 \pi}\right) \frac{4 \pi \varepsilon_{0}}{e^{2}} \end{aligned} $$
The above equation shows that $r$ is directly proportional to $n^{2}$
[CBSE Marking Scheme 2015]
Commonly Made Error
- Some students were unable to recall the correct equation.
i.e., $m v r=\frac{n h}{2 \pi}$ and $\frac{m v^{2}}{r}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{e^{2}}{r^{2}}$
Q. 11. The electron, in a hydrogen atom, is in its second excited state.
Calculate the wavelength of the lines in the Lyman series, that can be emitted through the permissible transitions of this electron.
(Given the value of Rydberg constant, $R=1.1 \times$ $10^{7} \mathrm{~m}^{-1}$ )
A [Delhi Comptt., I, II, III 2016]
Show Answer
Solution:
Ans. For second excited state, $n=3$
Hence two possible transition of the Lyman series : $3 \rightarrow 1$ and $2 \rightarrow 1$.
Wavelength for transition $3 \rightarrow 1, n_{f}=1, n_{i}=3$
$$ \begin{aligned} \frac{1}{\lambda} & =\mathrm{R}\left(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right) \ \frac{1}{\lambda} & =1.1 \times 10^{7}\left(\frac{1}{1}-\frac{1}{9}\right) \ & =1.1 \times 10^{7}\left(\frac{8}{9}\right) \ \Rightarrow \quad \lambda & =\frac{9}{8 \times 1.1 \times 10^{-7}} \ & =1.023 \times 10^{-7} \ & =102.3 \mathrm{~nm} \end{aligned} $$
For transition $2 \rightarrow 1, n_{f}=1, n_{i}=2$
$$ \frac{1}{\lambda}=1.1 \times 10^{7}\left(1-\frac{1}{4}\right) $$
$\Rightarrow$
$$ \lambda=212 \mathrm{~nm} $$
[CBSE Marking Scheme 2016]
Commonly Made Error
- Many students couldn’t remember that in Lyman series. $n_{f}=1$
Answering Tips
- Students should learn either with energy level diagram, or by drawing 7-8 orbits that how various series of lines are formed in the hydrogen spectrum.
