Physics MCQs for NEET — Practice Questions with Answers

(c) For Paschen series $\overline{\mathrm{v}}=\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left[\frac{1}{3^2}-\frac{1}{{\mathrm{n}}^2}\right]$ ; n=4, 5, 6....
For first member of Paschen series n = 4
$\frac{1}{{\mathrm{\lambda }}_1}=\mathrm{R}\left[\frac{1}{3^2}-\frac{1}{4^2}\right]\Rightarrow \frac{1}{{\mathrm{\lambda }}_1}=\frac{7\mathrm{R}}{144}$

$\Rightarrow \mathrm{R}=\frac{144}{7{\mathrm{\lambda }}_1}=\frac{144}{7\times 18800\times {10}^{-10}}=1.1\times {10}^{-7}$

For shortest wave length n = $\infty$
So

$$\begin{gathered} \frac{1}{\mathrm{\lambda }}=\mathrm{R}\left[\frac{1}{3^2}-\frac{1}{{\infty }^2}\right]=\frac{\mathrm{R}}{9} \\ \Rightarrow \mathrm{\lambda }=\frac{9}{\mathrm{R}}=\frac{9}{1.1\times {10}^{-7}}=8.225\times {10}^{-7} \mathrm{m} =8225 \stackrel{0}{\mathrm{A}} \end{gathered}$$

(d) For Lyman series $\frac{1}{{\mathrm{\lambda }}_{\max }}=\mathrm{R}\left[\frac{1}{1^2}-\frac{1}{2^2}\right]=\frac{3}{4}\mathrm{R}$  and 

$\frac{1}{{\mathrm{\lambda }}_{\min }}=\mathrm{R}\left[\frac{1}{1^2}-\frac{1}{{\infty }^2}\right]=\frac{\mathrm{R}}{1}\Rightarrow \frac{{\mathrm{\lambda }}_{\max }}{{\mathrm{\lambda }}_{\min }}=\frac{4}{3}$

(c) $\mathrm{T}\propto {\mathrm{n}}^3\Rightarrow \frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}=\frac{2^3}{1^3}=\frac{8}{1}$

(a) For Bracket series $\frac{1}{{\mathrm{\lambda }}_{\max }}=\mathrm{R}\left[\frac{1}{4^2}-\frac{1}{5^2}\right]=\frac{9}{25\times 16}\mathrm{R}$
and $\frac{1}{{\mathrm{\lambda }}_{\min }}=\mathrm{R}\left[\frac{1}{4^2}-\frac{1}{{\infty }^2}\right]=\frac{\mathrm{R}}{16}\Rightarrow \frac{{\mathrm{\lambda }}_{\max }}{{\mathrm{\lambda }}_{\min }}=\frac{25}{9}$

(a) For hydrogen and hydrogen like atoms ${\mathrm{E}}_{\mathrm{n}}=-13.6\frac{{\mathrm{z}}^2}{{\mathrm{n}}^2} \mathrm{eV}$

${\mathrm{U}}_{\mathrm{n}}=2{\mathrm{E}}_{\mathrm{n}}=-27.2 \frac{{\mathrm{z}}^2}{{\mathrm{n}}^2} \mathrm{eV}$and ${\mathrm{K}}_{\mathrm{n}}=\left|{\mathrm{E}}_{\mathrm{n}}\right|=13.6 \frac{{\mathrm{z}}^2}{{\mathrm{n}}^2} \mathrm{eV}$  

From these three relations we can see that as n decreases, ${\mathrm{K}}_{\mathrm{n}}$ will increase but ${\mathrm{E}}_{\mathrm{n}}$ and ${\mathrm{U}}_{\mathrm{n}}$ will decreases.

(a) $\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right]\Leftarrow \frac{{\mathrm{\lambda }}_{\min }}{{\mathrm{\lambda }}_{\max }}=\frac{{\displaystyle \left[\frac{{\displaystyle 1}}{{\displaystyle 2^2}}-\frac{{\displaystyle 1}}{{\displaystyle 3^2}}\right]}}{\left[\frac{1}{2^2}-\frac{1}{{\infty }^2}\right]}=\frac{5}{9}$

(b) It shows that atoms has a central positively charged nucleus that is surrounded by electrons.

(b) Maximum number of spectral lines are observed in Layman series.

(b) ${\mathrm{r}}_{\mathrm{n}}\propto {\mathrm{n}}^2\Rightarrow \frac{{\mathrm{r}}_3}{{\mathrm{r}}_1}=\frac{3^2}{1}\Rightarrow {\mathrm{r}}_3=9{\mathrm{r}}_1=9\times 0.53=4.77$Å

(d) For first line in Lyman series ${\mathrm{\lambda }}_{{\mathrm{L}}_1}=\frac{4}{3\mathrm{R}}$     ..... (i)
For first line in Balmer series ${\mathrm{\lambda }}_{{\mathrm{B}}_1}=\frac{36}{5\mathrm{R}}$          ..... (ii)
From equation (i) and (ii)

$\frac{{\mathrm{\lambda }}_{{\mathrm{B}}_1}}{{\mathrm{\lambda }}_{{\mathrm{L}}_1}}=\frac{27}{5}\Rightarrow {\mathrm{\lambda }}_{{\mathrm{B}}_1}=\frac{{\displaystyle 27}}{{\displaystyle 5}}{\mathrm{\lambda }}_{{\mathrm{L}}_1}\Rightarrow {\mathrm{\lambda }}_{{\mathrm{B}}_1}=\frac{{\displaystyle 27}}{{\displaystyle 5}}\mathrm{\lambda }$