Physics Test - 1

Required ionization energy

$=\frac{hc}{\lambda }=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{1.54\times {10}^{-10}}J$

$=12.9\times {10}^{-19} J$

Hence, the correct option is (b).

The radius of the charged particle moving in the circle is given by

\(r=\frac{m v}{q B}=\frac{p}{q B}\)

As the charge particles enter with same kinetic energy \(K=\frac{p^{2}}{2 m} \Rightarrow p=\sqrt{2 m K}\)

\(r=\frac{\sqrt{2 m K}}{q B} \Rightarrow r \propto \frac{\sqrt{m}}{q}\)

\(\mathrm{m}_{\alpha}=4 \mathrm{m}_{\mathrm{p}}\) and \(\mathrm{q}_{\alpha}=2 \mathrm{q}_{\mathrm{p}} \quad \mathrm{m}_{\mathrm{d}}=2 \mathrm{m}_{\mathrm{p}}\) and \(\mathrm{q}_{\mathrm{d}}=\mathrm{q}_{\mathrm{p}}\)

Hence \(r_{p} \propto \frac{\sqrt{m_{p}}}{q_{p}}\)

\(r_{d} \propto \frac{\sqrt{m_{d}}}{q_{d}}=\frac{\sqrt{2 m_{p}}}{q_{p}} \Rightarrow r_{d}=\sqrt{2} r_{p} \quad ; \quad r_{\alpha} \propto \frac{\sqrt{m_{\alpha}}}{q_{\alpha}}=\frac{\sqrt{4 m_{p}}}{2 q_{p}} \Rightarrow r_{\alpha}=r_{p}\)

\(r_{p}:r_{d}: r_{\alpha} =1: \sqrt{2}: 1\)

Number of emission spectral lines, $N=\frac{n\left(n-1\right)}{2}$

$\therefore 3=\frac{n_1\left(n_1-1\right)}{2}$, in the first case.

$n_1^2-n_1-6=0$ or $\left(n_1-3\right)\left(n_1+2\right)=0$

Take a positive root.

$\therefore n_1=3$

Again, $6=\frac{n_2\left(n_2-1\right)}{2}$, in second case.

$n_2^2-n_2-12=0$ or $\left(n_2-4\right)\left(n_2+3\right)=0$

Take positive root, or $n_2=4$

Now velocity of electron $v=\frac{2{\mathrm{\pi KZe}}^2}{nh}$

$\therefore \frac{v_1}{v_2}=\frac{n_2}{n_1}=\frac{4}{3}$

Let an electron (mass $m$), is revolving around the nucleus of a hydrogen atom in an orbit of radius $r$ with velocity $v$. Electron gets centripetal force to revolve, from the electrostatic attraction force between electron and nucleus.

$\frac{m{v}^2}{r}=\frac{e\times e}{4\mathrm{\pi }{\in }_0{\mathrm{r}}^2}$ (for $H$- atom, $Z=1$ hence charge on nucleus $=e$)

$m{v}^2=\frac{e^2}{4\mathrm{\pi }{\in }_0\mathrm{r}}$...(1)

According to Bohr's model, the angular momentum of the electron,

$mvr=\frac{nh}{2\mathrm{\pi }}$...(2)

Now, squaring equation (2) and dividing by equation (1), we get

$\frac{m{v}^2{r}^2}{m{v}^2}=\frac{n^2{h}^{2}4\mathrm{\pi }{\in }_0\mathrm{r}}{4{\mathrm{\pi }}^2{\mathrm{e}}^2}$

$r=\frac{n^2{h}^2{\in }_0}{{\mathrm{\pi me}}^2}$

Which gives $r\propto n^2$

According to the question, the Ionisation potential of a hydrogen atom is $13.6eV$.

Therefore, the change in the energy involved in removing the electron from $n=2$$\mathrm{}$ is given by the formula 

$E\left(n\right)=\frac{-13.6}{n^2}eV/atom$

However, in this case the electron is completely removed from  $n=2$$\mathrm{}$, therefore

$\mathrm{\Delta }E=E\left(\mathrm{\infty }\right)-E\left(2\right)$

$\mathrm{\Delta }E=0-\left(\frac{-13.6}{2^2}\right)$

$\mathrm{\Delta }E=\frac{+13.6}{4}$$$

$\mathrm{\Delta }E=3.4\mathrm{eV}$

As we know,

$\Delta E=13.6Z^2\times \left(\frac{1}{2^2}-\frac{1}{n^2}\right)$

For the transition from excited state $n$ to first excited state i.e. $n_1= 2$,

$(10.20+17.00)\mathrm{eV}=13.6Z^2\times (\frac{1}{2^2}-\frac{1}{n^2})-\text{equation}\left(1\right)$

For the transition from excited state $n$ to second excited state i.e. $n_1= 3$,

$(4.24+5.95)\mathrm{eV}=13.6Z^2\times \left(\frac{1}{3^2}-\frac{1}{n^2}\right)-\text{ equation }\left(2\right)$

On dividing equation $1$ and $2$, 

$\begin{array}{l}1.18=\frac{n^2-4}{n^2-9}\\ 0.18n^2=6.6\\ n^2=36\\ n=6\end{array}$

We know that, energy expression of Bohrs theory

$E=\frac{-13.6}{n^2}eV$

Inversely proportional to $n^2$