Physics M.C.Q (1 Marks)

Statement $II$ is wrong as atom of most of the elements are stable and emit characteristic spectrum. But this statement is not true for every atom.

$\therefore \lambda \propto \frac{1}{\Delta E}$

$(\Delta E)_{6-2}>(\Delta E)_{5-2}>(\Delta E)_{4-2}>(\Delta E)_{3-2}$

$\lambda_{6-2}<\lambda_{5-2}<\lambda_{4-2}<\lambda_{3-2}$

$A-III, B-IV, C-II, D-I$

$\because \frac{1}{\lambda}= Rz ^2\left[\frac{1}{2^2}-\frac{1}{\infty^2}\right]$

$\frac{1}{\lambda}=\frac{ R }{4}$

Shortest wavelength is Bracket series when transition of $e^{-}$from $\infty$ to $n=4$

$\frac{1}{\lambda^{\prime}}= R (1)^2\left[\frac{1}{4^2}-\frac{1}{\infty^2}\right] \Rightarrow \frac{1}{\lambda^{\prime}}=\frac{ R }{16}$

Eq. $(1)$/Eq.$(2)$

$\frac{\lambda^{\prime}}{\lambda}=\frac{R}{4} \times \frac{16}{R} \Rightarrow \lambda^{\prime}=4 \lambda$

$r _{ n }=0.53 \times \frac{ n ^2}{ Z } \mathring A$

So, radius of third orbit

$r_3=0.53 \times \frac{(3)^2}{(1)} \mathring A=4.77 \mathring A$

$\lambda_{\min }=\frac{ hC }{ eV }$

hence $\lambda_{\min } \propto \frac{1}{V}$

So Ans. $\left(\frac{1}{ V }\right)$

$\frac{ R _1( n =2)}{ R _2( n =4)}=\frac{2^2}{4^2}=\frac{1}{4}=0.25$

$T _{1}=-13.6 \frac{ z ^{2}}{ n ^{2}}=-\frac{13.6}{4} eV$

Second excited state $\Rightarrow n =3$

$T _{2}=-13.6 \frac{ z ^{2}}{ n ^{2}}=-\frac{13.6}{9} eV$

$T _{1}: T _{2}=\frac{1}{4}: \frac{1}{9}=9: 4$

$\mathrm{KE}=-\mathrm{T.E} \quad \mathrm{PE}=2 \mathrm{TE}$

$\Rightarrow \mathrm{KE}=+3.4 \mathrm{eV} \quad \Rightarrow \mathrm{PE}=-6.8 \mathrm{eV}$

$m_m=207 m_e$

When all other quantities remain uncharged,

$r \propto \frac{n^2}{m}$ and $E \propto \frac{m}{n^2}$

$\therefore r_m=\frac{r_e m_e}{m_m}=\frac{0.53 \times 10^3}{207}=2.56 \times 10^{-13} m$

$\text { and } E_m=\frac{E_e m_m}{m_e}=-(13.6 \times 207)=-2.8\; keV$

So, Kinetic energy : Total energy $=1:-\,1$

$\frac{1}{\lambda_{0}}= R \left(1-\frac{1}{4}\right) \Rightarrow \frac{1}{\lambda_{0}}=\frac{3}{4} R \ldots$

shortest wavelength in its infrared region

$\frac{1}{\lambda^{\prime}}= R \left(\frac{1}{9}\right) R \quad \frac{\lambda^{\prime}}{\lambda_{0}}=\frac{27}{4}$

$\frac{1}{\lambda_{B}}=R c\left(\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right)=\frac{R c}{4}$

The wavelength of last line of Lyman series

$\frac{1}{{{\lambda _L}}} = Rc\left( {\frac{1}{{{1^2}}} - \frac{1}{{{\infty ^2}}}} \right) = Rc$

$\therefore \quad \frac{{{l_B}}}{{{\lambda _L}}} = \frac{4}{1} = 4$

$r_{0}=\frac{Z e^{2}}{\pi \varepsilon_{0} m v^{2}} $ $\therefore r_{0} \propto \frac{1}{m}$

so, $\frac{1}{\lambda}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{5 R}{36}$

and $\frac{1}{\lambda^{\prime}}=R\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)=\frac{7 R}{144}$

$\therefore \lambda^{\prime}=\frac{144}{7} \times \frac{5 \lambda}{36}=\frac{20 \lambda}{7}$

The wave number of the last line of the Balmer series in hydrogen spectrum is given by

$\frac{1}{\lambda } = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{\infty ^2}}}} \right) = $ $\frac{R}{4} = \frac{{{{10}^7}}}{4} = 0.25 \times {10^7}{{\text{m}}^{ - 1}}$

$(i)$ Most of the space inside the atom is empty because most of the -particles passed through the gold foil without getting deflected.

$(ii)$ Very few particles were deflected from their path, indicating that the positive charge of the atom occupies very little space.

$(iii)$ A very small fraction of $\alpha$-particles were deflected by $1800$,indicating that all the positive charge and mass of the gold atom were concentrated in a very small volume within the atom.

From the data, he also calculated that the radius of the nucleus is about 105 times less than the radius of the atom.

On the basis of his experiment, Rutherford put forward the nuclear model of an atom, which had the following features:

$(i)$ There is a positively charged centre in an atom called the nucleus. Nearly all the mass of an atom resides in the nucleus.

$(ii)$ The electrons revolve around the nucleus in circular paths.

$(iii)$ The size of the nucleus is very small as compared to the size of the atom

$400 \times {10^3} \times 1.6 \times {10^{ - 19}} = 9 \times {10^9}\frac{{(ze)\,(2e)}}{r}$

$ \Rightarrow 6.4 \times {10^{ - 14}}$$ = \frac{{9 \times {{10}^9} \times (82 \times 1.6 \times {{10}^{ - 19}}) \times (2 \times 1.6 \times {{10}^{ - 19}})}}{r}$

$ \Rightarrow r = 5.9 \times {10^{ - 13}}m = 0.59\,pm$.

$\Delta \mathrm{K} \cdot \mathrm{E} \cdot=\Delta \mathrm{P} \cdot \mathrm{E}$

$4 \times 10^{6}=\frac{9 \times 10^{9} \times 2 \times 92 \times 1.6 \times 10^{-19}}{r}$

$r=662.4 \times 10^{-16} \mathrm{\,m}$

$=6.62 \times 10^{-14} \mathrm{\,m} \Rightarrow 6.62 \times 10^{-12} \mathrm{\,cm}$

$N(\theta) \propto \frac{Z^2}{\sin ^4\left(\frac{\theta}{2}\right) E^2}$

$\text { or } N (\theta) \propto \frac{1}{\sin ^4\left(\frac{\theta}{2}\right)}$

$\theta=60^{\circ}$ and $90^{\circ}$

$\therefore \frac{N(90)}{N(60)} \propto \frac{\sin ^4 30}{\sin ^4 45}$

$N\left(90^{\circ}\right)=x$

$N\left(60^{\circ}\right)=4 x$

==> $2\pi r = n\,\left( {\frac{h}{{mv}}} \right) = n\lambda $ for $n = 1$, $\lambda = 2\pi r$

==> $\frac{{z{e^2}}}{{{r^2}}} = \frac{{m{v^2}}}{r}$ ==> $m{v^2} = \frac{{z{e^2}}}{r}$

$\therefore$ $ K.E.$ $ = \frac{1}{2}m{v^2} = \frac{{z{e^2}}}{{2r}}$

$K.E.$ $ = - \frac{1}{2}({\rm{P}}{\rm{.E}}{\rm{.}}) = \frac{{{e^2}}}{{8\pi {\varepsilon _0}r}}$

==> $\frac{{21.2 \times {{10}^{ - 11}}}}{{5.3 \times {{10}^{ - 11}}}} = {\left( {\frac{n}{1}} \right)^2}$ ==> ${n^2} = 4$ ==>$ n = 2$

Minimum excitation energy corresponds to excitation from $n = 1$ to $n = 2$

Minimum excitation energy in hydrogen atom $ = - 3.4 - ( - 13.6) = + 10.2\,eV$ 

so minimum excitation potential $= 10.2\, eV.$

$\Delta L = {L_2} - {L_1} = \frac{{{n_2}h}}{{2\pi }} - \frac{{{n_1}h}}{{2\pi }}$$ \Rightarrow \Delta L = \frac{h}{{2\pi }}({n_2} - {n_1})$ 

$ = \frac{{6.6 \times {{10}^{ - 34}}}}{{2 \times 3.14}}(5 - 4)$$ = 1.05 \times {10^{ - 34}}J{\rm{ - }}S$

So, $\frac{m v^2}{r}=\frac{k e^2}{r^2}$

$\Rightarrow m v^2=\frac{k e^2}{r}-\text { (1) }$

Again,

$\Rightarrow v=\frac{n h}{2 \pi m r}$

from equation $- (1)$

$\Rightarrow m\left(\frac{n h}{2 \pi m h}\right)^2=\frac{k e^2}{r}$

$\therefore r=\frac{n^2 h^2}{4 \pi^2 k m e^2}-\text { (2) }$

Potential energy, $E_p=-\frac{k_e^2}{r}=-\frac{4 \pi^2 k^2 m e^4}{n^2 h^2}$

Hence total energy of the electron in the $n-$the orbit, $E=E_K+E_p$

$=-\frac{2 \pi^2 k^2 m e^4}{n^2 h^2}$

The electron in a hydrogen atom travels around the nucleus in a circular orbit. The nucleus is of infinite mass and is rest.

Electrons in a quantized orbit will not radiate energy. Mass of electron remains constant.

and energy $E \propto m$, where $m$ is the mass of the electron.

Hence energy of hypothetical atom

${E_0} = 2 \times ( - 13.6\,eV) = - 27.2\,eV$ and radius ${r_0} = \frac{{{a_0}}}{2}$

Velocity $v = \frac{{Z{e^2}}}{{2{\varepsilon _0}nh}}$

and energy $E = - \frac{{m{Z^2}{e^4}}}{{8\varepsilon _0^2{n^2}{h^2}}}$

Now, it is clear from above expressions $R\cdot v \propto n$

==> $\frac{{{A_n}}}{{{A_1}}} = {\left( {\frac{{{r_n}}}{{{r_1}}}} \right)^2} = {\left( {\frac{n}{1}} \right)^4}$                      $(\because {r_n} \propto {n^2})$

Taking loge both the side ${\log _e}\frac{{{A_n}}}{{{A_1}}} = 4\,{\log _e}(n)$

Comparing it with $y = mx + c$, graph $(4)$ is correct.
 

$= 10.6 Å.$

${r_n} = 0.53\frac{{{n^2}}}{Z}\mathring A $

Here for $n =1, E_1 = -54.4\, eV$

Therefore $ - \,54.4 = - 13.6\frac{{{Z^2}}}{{{1^1}}}$ ==> $Z = 2$

Hence radius of first Bohr orbit $r = \frac{{0.53\,{{(1)}^2}}}{2} = 0.265\mathring A$

==> $E_{ionisation}= 13.6 Z^2$
==> $\frac{{{{({E_{ion}})}_H}}}{{{{({E_{ion}})}_{Li}}}} = {\left( {\frac{{{Z_H}}}{{{Z_{Li}}}}} \right)^2} = {\left( {\frac{1}{3}} \right)^2} = \frac{1}{9}$

$n_{r}=\frac{\lambda_{r}}{\lambda_{b}} n_{b}$

As $\lambda _r > \lambda _b$

$\therefore \ \boxed {n_r > n_b}$

$a_{0}=\frac{h^{2} \varepsilon_{0}}{\pi m e^{2}}$

Therefore, the Bohr radius is inversely proportional to the mass of the orbiting particle.

$\therefore \frac{a_{0 e}}{a_{0 m}}=\frac{m_{m}}{m_{e}}$

$\therefore a_{0 m}=\frac{m_{e} a_{0 e}}{m_{m}}=\frac{m_{e} a_{0 e}}{207 m_{e}}$

$\therefore a_{0 m}=\frac{a_{0 e}}{207}=\frac{5.29 \times 10^{-11}}{207}=2.56 \times 10^{-13} \mathrm{m}$

$\therefore a_{0 m}=2.56 \times 10^{-3} A$

$\frac{1}{\lambda}=R z^{2}\left[\frac{1}{n_{2}^{2}}-\frac{1}{n_{1}^{2}}\right] \Rightarrow \frac{1}{\lambda} \propto z^{2}$

$\lambda$ is shortest when $1 / \lambda$ is i.e., when $z$ is big $z$ is highest for lithum.

${m v r=\frac{n h}{2 \pi}}$        .......$(2)$

$n=1$       .........$(3)$

$\lambda=\frac{h}{p}$

${\rm{L}} = \frac{{{\rm{n}}\lambda }}{2}\,\,\,\,\,\,\lambda  = \frac{{2{\rm{L}}}}{n}$

$\mathrm{p}=\frac{\mathrm{h}}{\lambda}=\mathrm{h} \frac{\mathrm{h}}{2 \mathrm{L}} \mathrm{p}=\frac{\mathrm{h}}{\lambda}=\frac{\mathrm{nh}}{2 \mathrm{L}}$

for $\mathrm{H}, \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{1}-\frac{1}{4}\right],$ for $\mathrm{H}_{\mathrm{e}}, \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\left(\frac{\mathrm{n}_{\mathrm{f}}}{\mathrm{z}}\right)^{2}}-\frac{1}{\left(\frac{\mathrm{n}_{\mathrm{i}}}{\mathrm{z}}\right)^{2}}\right]$

$\Rightarrow \frac{\mathrm{n}_{\mathrm{f}}}{2}=1 \quad \Rightarrow \mathrm{n}_{\mathrm{f}}=2 | \frac{\mathrm{n}_{\mathrm{i}}}{\mathrm{z}}=2 \Rightarrow \mathrm{n}_{\mathrm{i}}=4$

$\frac{1}{\lambda}=\frac{24}{25} \mathrm{R}$

$\mathrm{P}=\mathrm{mv}=\frac{\mathrm{h}}{\lambda}$

$v=\frac{h}{m \lambda}=\frac{24 h R}{25 m}$