NEET Practice Questions (MCQs) with Answers & Solutions

(a) The concept of stationary orbits was proposed by Neil Bohr.

(a) 

(b) $\mathrm{T}=\frac{2\mathrm{\pi r}}{\mathrm{v}}$  ; r = radius of nth orbit= $\frac{{\mathrm{n}}^2{\mathrm{h}}^2}{{\mathrm{\pi mZe}}^2}$

v = speed of ${\mathrm{e}}^{-}$ in nth orbit = $\frac{{\mathrm{ze}}^2}{2{\mathrm{\epsilon }}_0\mathrm{nh}}$

$\therefore \mathrm{T}=\frac{4{\mathrm{\epsilon }}_0^2{\mathrm{n}}^3{\mathrm{h}}^3}{{\mathrm{mZ}}^2{\mathrm{e}}^4}\Rightarrow \mathrm{T}\propto \frac{{\mathrm{n}}^3}{{\mathrm{Z}}^2}$

(b) For Lyman series

$\overline{\mathrm{v}}=\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left(\frac{1}{1^2}-\frac{1}{{\mathrm{n}}^2}\right)$ here n =2, 3, 4, 5......

For first line 

$\overline{\mathrm{v}}=\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \Rightarrow \overline{\mathrm{v}}=\mathrm{R}\left(1-\frac{1}{4}\right)=\frac{3\mathrm{R}}{4}$

(b) Energy $\mathrm{E}=\mathrm{K}\left[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right]$ (K = constant)
${\mathrm{n}}_1$ = 2 and ${\mathrm{n}}_2$ = 3, so $\mathrm{E}=\mathrm{K}\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=\mathrm{K}\left[\frac{5}{36}\right]$
For removing an electron ${\mathrm{n}}_1$ = 1 to ${\mathrm{n}}_2$ = $\infty$
Energy ${\mathrm{E}}_1=\mathrm{K}\left[1\right]=\frac{36}{5}\mathrm{E}=7.2 \mathrm{E}$
$\therefore$ Ionization energy = 7.2 E

(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.