Physics MCQs for NEET — Practice Questions with Answers

$$\begin{gathered} E_n=\frac{-13.6}{n^2}eV; (given E_n=-3.4 eV) \\ \therefore -3.4=\frac{-13.6}{n^2}; n=2 \end{gathered}$$

Therefore, angular momentum of electron in 2nd orbit is required which is given by 

$$\begin{gathered} L_n=mv{r}_n=\frac{nh}{2\mathrm{\pi }}; n=2 \\ {L}_2=\frac{2\times 6.6\times {10}^{-34}}{2\times 3.14}=2.11\times {10}^{-34} Js \end{gathered}$$

The radius of nth Bohr's orbit of hydrogen and hydrogen like atom

$$\begin{gathered} r_n=\frac{{\epsilon }_o{n}^2{h}^2}{\pi m{e}^{2}Z} \\ \therefore r_n=\frac{n^2{a}_0}{Z} or r_n \alpha \frac{n^2}{Z} \end{gathered}$$

For ground state, n = 1

Atomic number, Z = 1

For first excited state, n = 2 

$$\begin{gathered} \therefore \frac{r_2}{r_1}={\left(\frac{2}{1}\right)}^2=4 \\ or r_2=4r_1 \end{gathered}$$

Therefore, radius of first excited state is 4 times than that of ground state radius in H-atom.

Energy, W = qV, ionisation energy in eV = Ionisation Potential in Volt

$\therefore$24.6 eV energy is required to ionise helium atom.

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

For first line of Lyman series $n_1$= 1 and $n_2$= 3

For first line of Balmer series $n_1$= 2 and  $n_2$= 3

So, $\frac{{\lambda }_{Lyman}}{{\lambda }_{Balmer}}=\frac{5}{27}$

$\left(E_{ion}\right)Na=Z^2\left(E_{ion}\right)H={\left(11\right)}^2\times 13.6 eV$

Ionisation energy is defined as the energy required to knock an electron completely out of an isolated gaseous atom. When electron is raised to the orbit n = $\infty$, it will be completely out of the atom. Ionisation energy of helium,

$$\begin{gathered} E=\frac{2{\mathrm{\pi }}^2{\mathrm{mZ}}^2{\mathrm{k}}^2{\mathrm{e}}^4}{h^2}\left[\frac{1}{1^2}-\frac{1}{\infty }\right] \\ =\frac{2{\mathrm{\pi }}^2\mathrm{m}{\left(2\right)}^2{\mathrm{k}}^2{\mathrm{e}}^4}{h^2}\left[\frac{1}{1^2}-\frac{1}{\infty }\right] \\ Ionisation energy for hydrogen atom \\ =13.6 eV=\frac{2{\mathrm{\pi }}^2{\mathrm{mk}}^2{\mathrm{e}}^2}{h^2}\times 4\times \left[\frac{1}{1^2}-\frac{1}{{\infty }^2}\right] \\ \Rightarrow 4\times 13.6=54.4 eV \end{gathered}$$

Energy of $n^{th}$ state in hydrogen is same as energy of $3n^{th}$ state in $L{i}^{++}$.

$\therefore$ 3$\to$1 transition in H would give same energy as the 

3 $\times$3$\to$1$\times$3 transition in $L{i}^{++}$

$$\begin{gathered} \frac{1}{\lambda }=R\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right) \\ C=\nu \lambda ; \frac{1}{\lambda }=\frac{\nu }{C} \\ \frac{\nu }{C}=R\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right) \\ \nu =CR\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{5}{36}CR \end{gathered}$$

$$\begin{gathered} E_n=-\frac{13.6}{n^2}eV \\ =\frac{13.6}{5^2}eV \\ =-0.54 eV \end{gathered}$$

$$\begin{gathered} For Brackett series, \overline{v}=R\left(\frac{1}{4^2}-\frac{1}{n^2}\right) \\ \frac{9}{400}R=R\left(\frac{1}{16}-\frac{1}{n^2}\right) or \frac{9}{400}=\frac{1}{16}-\frac{1}{n^2} \\ \frac{1}{n^2}=\frac{1}{16}-\frac{9}{400}=\frac{16}{400} \\ {n}^2=\frac{400}{16} or n=\frac{20}{4}=5 \end{gathered}$$