NEET Practice Questions (MCQs) with Answers & Solutions

According to Bohr's hypothesis, electron can revolve only in those orbits in which its angular momentum is an integral multiple of $\frac{h}{2\mathrm{\pi }}$, where h is Plank's constant.  In these orbits, angular momentum of electron can gave magnitude as $\frac{h}{2\mathrm{\pi }}, \frac{2h}{2\mathrm{\pi }}, \frac{3h}{2\mathrm{\pi }}....$etc, but never as $\frac{1.5 h}{2\mathrm{\pi }}, \frac{2.5 h}{2\mathrm{\pi }}, \frac{3.5 h}{2\mathrm{\pi }}....$etc.  This is called the quantisation of angular momentum.

When a hydrogen atom is raised from the ground state to excited state Potential energy increases and KE decreases, because

$$\begin{gathered} KE=\frac{1}{2}m{v}^2 \\ where \frac{m{v}^2}{r_n}=\frac{Z{e}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}_{\mathrm{n}}^2} \\ m{v}^2=\frac{Z{e}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}_{\mathrm{n}}} \\ \therefore KE=\frac{1}{2}\left(\frac{Z{e}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}_{\mathrm{n}}}\right)=\frac{Z{e}^2}{8{\mathrm{\pi \epsilon }}_0{\mathrm{r}}_{\mathrm{n}}} \\ PE=-\frac{Z{e}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}_{\mathrm{n}}} [-ve of twice the KE] \end{gathered}$$

When the electron jumps from n = 1 to n = 2

since $r_n \alpha n^2$ radius of orbit increases

$\therefore KE decreases$

Since PE is -ve, when $r_n$ increases, $\frac{Z{e}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}_{\mathrm{n}}}$ decreases - $\frac{Z{e}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}_{\mathrm{n}}}$ increases

So PE increases, KE decreases,

$\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For Paschen series, $n_1=3$ for extreme lines of Paschen series $n_2=4$ and $n_2=\infty$

$$\begin{gathered} \therefore \frac{1}{{\lambda }_1}=R\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=1.097\times {10}^7(0.0486) \\ {\lambda }_1=1.88\times {10}^{-6}m=1.88 \mu m \\ and \frac{1}{{\lambda }_2}=R\left(\frac{1}{3^2}-\frac{1}{{\infty }^2}\right)=1.097\times {10}^7\left(\frac{1}{9}\right) \\ {\lambda }_2=8.2\times {10}^{-7}m \\ =0.82\times {10}^{-6}m=0.82 \mu m \end{gathered}$$

$$\begin{gathered} As \frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \therefore \frac{1}{\lambda }=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right) \\ Multiply both sides by \lambda \\ 1=\lambda R\left(1-\frac{1}{n^2}\right) or \frac{1}{\lambda R}=1-\frac{1}{n^2} \\ or \frac{1}{n^2}=1-\frac{1}{\lambda R}=\frac{\lambda R-1}{\lambda R} \\ or n=\sqrt{\frac{\lambda R}{\lambda R-1}} \end{gathered}$$

The circumference of an orbit in an atom in terms of wavelength of wave associated with electron is given by 

$$\begin{gathered} 2{\mathrm{\pi r}}_{\mathrm{n}}=\mathrm{n\lambda } \\ [{\mathrm{r}}_{\mathrm{n}}=\mathrm{radius} \mathrm{of} \mathrm{any} \mathrm{n} \mathrm{orbit}] \end{gathered}$$

From Coulomb's attraction between the positive proton and negative electron = $\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{e^2}{r^2} [For neutral atom]$

Centripetal force has magnitude  $F=\frac{m{v}^2}{r}$

So for the revolving electrons

$$\begin{gathered} \frac{m{v}^2}{r}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{e^2}{r^2} \\ \Rightarrow v^2=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{e^2}{mr} \\ or v=\frac{e}{\sqrt{4{\mathrm{\pi \epsilon }}_0\mathrm{mr}}} \\ \mathrm{For} \mathrm{ground} \mathrm{state} \mathrm{of} \mathrm{H}-\mathrm{atom}, \mathrm{r}={\mathrm{a}}_0 \\ \therefore \mathrm{v}=\frac{\mathrm{e}}{\sqrt{4{\mathrm{\pi \epsilon }}_0{\mathrm{ma}}_0}} \end{gathered}$$

For,  second excited state n = 3

$\therefore$Energy needed to ionise H-atom from its second excited state 

$$\begin{gathered} E=\frac{2{\mathrm{\pi }}^2{\mathrm{mke}}^4}{h^2}\left(\frac{1}{3^2}-\frac{1}{\infty }\right) \\ or we can say that \\ E \alpha \frac{Z^2}{n^2} \\ \left(Z=atomic number\right) \\ (n=nth orbit) \\ So, E=\frac{13.6}{3^2}eV=1.51 eV \end{gathered}$$

$$\begin{gathered} \frac{1}{{\lambda }_a}=\frac{3R}{4}{(Z-1)}^2\Rightarrow (Z-1)=\sqrt{\frac{4}{3R{\lambda }_{\alpha }}}=\sqrt{\frac{4}{3\times 1.1\times {10}^7\times 1.8\times {10}^{-10}}} \\ =\frac{200}{3}\sqrt{\frac{5}{33}}=\frac{78}{3}=26 \\ \Rightarrow Z=27 \end{gathered}$$

$$\begin{gathered} \frac{1}{{\lambda }_{H_2}}=R{Z}_H^2\left[\frac{1}{4}-\frac{1}{9}\right]=R{\left(1\right)}^2\left[\frac{5}{36}\right] \\ \frac{1}{{\lambda }_{He}}=R{Z}_{He}^2\left[\frac{1}{4}-\frac{1}{16}\right]=R\left(4\right)\left[\frac{3}{16}\right] \end{gathered}$$

$$\begin{gathered} \frac{{\lambda }_{He}}{{\lambda }_{H_2}}=\frac{1}{4}\left[\frac{16}{3}\times \frac{5}{36}\right]=\frac{5}{27} \\ {\lambda }_{He}=\frac{5}{27}\times 6561=1215 \stackrel{0}{A} \end{gathered}$$

Given, $E_n=-\frac{13.6}{n^2}eV$

Energy of photon ejected when electron jumps from n = 3 to n = 2 state is given by 

$∆E=E_3-E_2$

Energy of third orbit

$$\begin{gathered} E_3=-\frac{13.6}{{\left(3\right)}^2}eV \\ =-\frac{13.6}{9}eV \\ Energy of second orbit \\ {E}_2=-\frac{13.6}{{\left(2\right)}^2}eV=-\frac{13.6}{4}eV \\ So, ∆E=E_3-E_2=-\frac{13.6}{9}-\left(-\frac{13.6}{4}\right) \\ =1.9 eV \left(approximately\right) \end{gathered}$$