Physics MCQs for NEET — Practice Questions with Answers

Ionisation energy corresponding to ionisation potential ($E_1=-13.6 eV$)

Photon energy incident ($∆E$) = 12.1 eV

So, the energy of electron in excited state ($E_2$) is given by

$$\begin{gathered} E_2-E_1=∆E \\ \Rightarrow E_2=∆E+E_1 \\ \Rightarrow E_2=-13.6+12.1 \\ \Rightarrow E_2=-1.5 eV \\ i.e. E_2=-\frac{13.6}{n^2}eV \\ -1.5=\frac{-13.6}{n^2} \\ \Rightarrow n^2=\frac{-13.6}{-1.5}\approx 9 \\ \therefore n=3 \\ i.e energy of electron in excited state corresponds to third orbit. \\ The possible spectral lines is given by \\ \frac{n(n-1)}{2} \\ \Rightarrow \frac{3(3-1)}{2} \\ \Rightarrow 3 \end{gathered}$$

$$\begin{gathered} \frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right); C=c\lambda , \frac{1}{\lambda }=\frac{\nu }{C} \\ \frac{\nu }{C}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right); \nu =CR\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \end{gathered}$$

For series limit n, Balmer series $n_1=2, n_2=\infty$

$\therefore \nu =RC\left(\frac{1}{2^2}-\frac{1}{{\infty }^2}\right)=\frac{RC}{4}$

$$\begin{gathered} m\nu r=\frac{nh}{2\mathrm{\pi }}\Rightarrow \nu =\frac{nh}{2\mathrm{\pi mr}} \\ \Rightarrow \frac{{\mathrm{\nu }}^2}{\mathrm{r}}=\frac{{\mathrm{n}}^2{\mathrm{h}}^2}{4{\mathrm{\pi }}^2{\mathrm{m}}^2{\mathrm{r}}^3} \end{gathered}$$

$\frac{1}{\lambda }=\frac{Z^{2}m{e}^4}{8{\epsilon }_0^{2}C{h}^3}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

Hydrogen atom Z = 1

Deutron Z = 1

Singly ionised He , Z = 2

Doubly ionised Li , Z = 3         (Gives shortest $\lambda$)

Visible light has wavelength ranging from 4000 $\stackrel{\circ }{A}$ to 7000 $\stackrel{\circ }{A}$ from violet to red, $\lambda$ red > $\lambda$ violet. Ultraviolet has shorter wavelength than violet (4000 $\stackrel{\circ }{A}$). Infra red has longer wavelength than red (7000 $\stackrel{\circ }{A}$). Therefore, choice D is correct, which is a false statement.

If $r_0$ is or distance of closest approach, 

$$\begin{gathered} KE=PE \\ E=\frac{1}{2}m{v}^2=\frac{\left(Ze\right) \left(2e\right)}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}} \\ \therefore \mathrm{r}=\frac{\left(\mathrm{Ze}\right)2\mathrm{e}}{4{\mathrm{\pi \epsilon }}_0\mathrm{E}}=\frac{9\times {10}^9\times 92\times 1.6\times {10}^{-19}\times 2\times 1.6\times {10}^{-19}}{5\times {10}^6\times 1.6\times {10}^{-19}} \\ \approx {10}^{-14} \mathrm{m} \end{gathered}$$

Ionisation energy = $Z^2 Rhc \left(\frac{1}{{\left(1\right)}^2}-\frac{1}{{(\infty )}^2}\right)$

 $$\begin{gathered} =Z^{2}Rhc={\left(3\right)}^2\times 13.6 eV \\ =122.4 eV \end{gathered}$$

Angular momentum $L=n\left(\frac{h}{2\mathrm{\pi }}\right)$

For this case n = 2, hence $L=2\times \frac{h}{2\mathrm{\pi }}=\frac{h}{\mathrm{\pi }}$

Total energy of electron for hydrogen like atom is given by

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

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