Physics MCQs for NEET — Practice Questions with Answers

(a) For Lyman series

${\mathrm{v}}_{\mathrm{Lyman}}=\frac{\mathrm{c}}{{\mathrm{\lambda }}_{\max }}=\mathrm{Rc}\left[\frac{1}{{\left(1\right)}^2}-\frac{1}{{\left(2\right)}^2}\right]=\frac{3\mathrm{RC}}{4}$

For Balmer series

${\mathrm{v}}_{Balmer}=\frac{\mathrm{c}}{{\mathrm{\lambda }}_{\max }}=\mathrm{Rc}\left[\frac{1}{{\left(2\right)}^2}-\frac{1}{{\left(3\right)}^2}\right]=\frac{5\mathrm{RC}}{36}$

$\therefore \frac{{\mathrm{v}}_{\mathrm{Lyman}}}{{\mathrm{v}}_{\mathrm{Balmer}}}=\frac{27}{5}$

The frequency of a photon emitted during a transition is inversely proportional to the wavelength. The transition from n=2 to n=1 in the hydrogen atom corresponds to the shortest wavelength in the Balmer series, which means it has the highest frequency.

(c) Wave number = $\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left(\frac{1}{{\mathrm{n}}_1^2}-\frac{{\displaystyle 1}}{{\displaystyle {\mathrm{n}}_2^2}}\right)$
For first Balmer line ${\mathrm{n}}_1$ = 2, ${\mathrm{n}}_2$=3
Wave number

$\therefore$ Wave number = $\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\mathrm{R}\left(\frac{9-4}{9\times 4}\right)=\frac{5\mathrm{R}}{36}$

The frequency of the emitted photon is inversely proportional to the square of the principal quantum number (n). The transition from n=4 to n=3 involves the smallest energy difference and hence the lowest frequency photon emission.

(a) $\mathrm{P}.\mathrm{E}. =-\frac{{\mathrm{ke}}^2}{\mathrm{r}}=-\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}}; \mathrm{K}.\mathrm{E}=-\frac{1}{2}\left(\mathrm{PE}\right)=\frac{{\mathrm{e}}^2}{8{\mathrm{\pi \epsilon }}_0\mathrm{r}}$

(c) $\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left(\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right)$
For first line of Lymen series ${\mathrm{n}}_1$ = 1 and ${\mathrm{n}}_2$ = 2
For first line of Balmer series ${\mathrm{n}}_1$ = 2 and ${\mathrm{n}}_2$ = 3
So, $\frac{{\mathrm{\lambda }}_{\mathrm{Lyman}}}{{\mathrm{\lambda }}_{\mathrm{Balmer}}}=\frac{5}{27}$

(d) ${\mathrm{v}}_{\mathrm{n}}\propto \frac{1}{\mathrm{n}}\Rightarrow \frac{{\mathrm{v}}_5}{{\mathrm{v}}_2}=\frac{2}{5}\Rightarrow {\mathrm{v}}_5=\frac{2}{5}{\mathrm{v}}_2=\frac{2}{5}\mathrm{v}$

(d) By using ${\mathrm{N}}_{\mathrm{E}}=\frac{\mathrm{n}(\mathrm{n}-1)}{2}\Rightarrow {\mathrm{N}}_{\mathrm{E}}=\frac{4(4-1)}{2}=6$

(d) Shortest wavelength comes from ${\mathrm{n}}_1=\infty$ to ${\mathrm{n}}_2=1$ and longest wavelength comes from ${\mathrm{n}}_1=6$ to ${\mathrm{n}}_2=5$ in the given case. Hence

$$\begin{gathered} \frac{1}{{\mathrm{\lambda }}_{\min }}=\mathrm{R}\left(\frac{1}{1^2}-\frac{1}{{\infty }^2}\right)=\mathrm{R} \\ \frac{{\displaystyle 1}}{{\displaystyle {\mathrm{\lambda }}_{\max }}}=\mathrm{R}\left(\frac{{\displaystyle 1}}{{\displaystyle 5^2}}-\frac{{\displaystyle 1}}{{\displaystyle 6^2}}\right)=\mathrm{R}\left(\frac{36-25}{25\times 36}\right)=\frac{11}{900}\mathrm{R} \end{gathered}$$

$\therefore \frac{{\mathrm{\lambda }}_{\max }}{{\mathrm{\lambda }}_{\min }}=\frac{900}{11}$

(b) As n increases P.E. increases and K.E. decreases.