NEET Practice Questions (MCQs) with Answers & Solutions

Total energy of electron for hydrogen like atom is given by

$$\begin{gathered} E_n=-\frac{13.6 Z^2}{n^2} \\ \therefore E_3=-\frac{13.6}{3^2}eV [Z=1, n=3] \\ =-1.51 eV \\ {E}_4=-\frac{13.6}{4^2} \\ =-0.85 eV \\ \therefore E_4-E_3=1.51-0.85 \\ =0.66 eV \end{gathered}$$

$$\begin{gathered} E_n=\frac{-13.6}{n^2}eV \\ For first excited state, n=2 \\ \therefore E_2=\frac{-13.6}{2^2}eV \\ For second excited state, n=3 \\ {E}_3=\frac{-13.6}{3^2}eV \\ \therefore \frac{E_2}{E_3}=\frac{9}{4} \end{gathered}$$

 (4)

The impact parameter is the perpendicular distance of the velocity vector of the alpha particle from the central line of the nucleus when the particle is far away from the nucleus of the atom.

Rutherford calculated analytically, the relation between the impact parameter b and scattering angle $\theta$, which is given by 

      b = $\frac{1}{4{\mathrm{\pi e}}_0}· \frac{Z{e}^{2}cot{\displaystyle \frac{\theta }{2}}}{E}$

Where, E = $\frac{1}{2}m{v}^2$ is the kinetic energy of the alpha particle, when it is far away from the atom.

According to problem,

      b = $\frac{1}{4{\mathrm{\pi e}}_0}· \frac{Z{e}^{2}cot{\displaystyle \frac{\theta }{2}}}{E}$ = 0

As given that b = 0

so,      cot$\frac{\theta }{2}$ = 0

$\therefore$           $\frac{\theta }{2}$= ${90}^{\circ }$

or $\theta$ = ${180}^{\circ }$

$v=\frac{c}{137}=\frac{3\times {10}^8}{137}=2\times {10}^6 m{s}^{-1}$

$$\begin{gathered} \lambda \left(K\alpha \right)=\frac{1216}{{(Z-1)}^2} \stackrel{0}{A} \\ or {(Z-1)}^2=676 \end{gathered}$$

Therefore, Z=27, the element is cobalt

The frequencies of the emitted photon in the paschen series are given by, $v=Rc\left(\frac{1}{3^2}-\frac{1}{n^2}\right)$

Where n = 4, 5, 6, ..............

The highest frequency corresponds to n = $\infty$

$$\begin{gathered} \therefore v_{highest}=\frac{Rc}{9} \\ =\frac{1.097\times {10}^7 m^{-1}\times 3\times {10}^8 m/s}{9} \\ =0.37\times {10}^{15} s^{-1} \\ =3.7\times {10}^{14} s^{-1} \\ =3.7\times {10}^{14} Hz \end{gathered}$$

$$\begin{gathered} \mathrm{After} \mathrm{removing} \mathrm{one} \mathrm{electron}, \mathrm{He} \mathrm{atom} \mathrm{becomes} \mathrm{Hydrogen} \mathrm{like} \mathrm{atom}. \\ \mathrm{Energy} \mathrm{of} \mathrm{hydrogen} \mathrm{atom}=-13.6\frac{{\mathrm{z}}^2}{{\mathrm{n}}^2}\mathrm{eV} \\ \mathrm{Energy} \mathrm{required} \mathrm{to} \mathrm{remove} 2\mathrm{nd} \mathrm{electron}=13.6\frac{2^2}{1^2}=54.4\mathrm{eV} \\ \mathrm{Total} \mathrm{energy} \mathrm{required}=54.4+24.6=79 \mathrm{eV} \end{gathered}$$

 $\frac{1}{\lambda }= R\left[\begin{array}{c}\frac{1}{{n_1}^2}- \frac{1}{{n_2}^2}\\ \end{array}\right]$ $\Rightarrow \frac{1}{{\lambda }_{3\to 2}} = R \left[\begin{array}{cc}\frac{1}{(2{)}^2}-& \frac{1}{(3{)}^2}\end{array}\right] = \frac{5R}{36}$

and $\frac{1}{{\lambda }_{4\to 2}} = R\left[\begin{array}{cc}\frac{1}{(2{)}^2}-& \frac{1}{(4{)}^2}\end{array}\right] = \frac{3R}{16}$

$\therefore$ $\frac{{\lambda }_{4\to 2}}{{\lambda }_{3\to 2}}= \frac{20}{27} \Rightarrow {\lambda }_{4\to 2}= \frac{20}{27}{\lambda }_{\circ }$

Energy levels of H-atoms are given by

$$\begin{gathered} E_n=-\frac{13.6 Z^2}{n^2}eV (Z=1) \\ \Rightarrow E_n=-13.6/n^2 eV \end{gathered}$$

Photons are emitted only when electron jumps from higher energy level (higher n- value) to lower energy level (lower n-value). So, alternative (1) and (3) are wrong. 

Energy difference from n = 2 to n = 1 level is 

$$\begin{gathered} ∆E_{2\to 1}=13.6\left(\frac{1}{1^2}-\frac{1}{2^2}\right)eV \\ =13.6\times \frac{3}{4}=10.2 eV \\ Energy difference from n=6 to n=2 level is \\ ∆E_{6\to 2}=13.6\left(\frac{1}{2^2}-\frac{1}{6^2}\right) \\ =13.6\times \left(\frac{1}{4}-\frac{1}{36}\right) \\ =13.6\times \frac{2}{6}=3.02 eV \end{gathered}$$

Thus, it is evident that difference is larger for n = 2 to n = 1 transition. Hence, maximum energy photons or shortest wavelength will be emitted during transition from n = 2 to n = 1.