Atoms and Nuclei

${{hc} \over \lambda } = {E_4} - {E_3}$

${{hc} \over \lambda } = - 13.68 \times {2^2}\left( {{1 \over {{4^2}}} - {1 \over {{3^2}}}} \right)$

${{hc} \over \lambda } = 13.6 \times {2^2}\left( {{1 \over 9} - {1 \over {16}}} \right)$

${{4.1356 \times {{15}^{ - 15}}\,eV.s \times 3 \times {{10}^8}\,m{s^{ - 1}}} \over 1} = 2.644$

$ \Rightarrow \lambda \cong 4.8\, * \,{10^{ - 7}}\,m$

To determine the ratio of the kinetic energy of the $n=2$ electron for the H atom to that of the He$^+$ ion, we need to consider the Bohr model of the atom. According to the Bohr model, the kinetic energy (KE) of an electron in a hydrogen-like ion in the nth orbit can be given by:

$ KE_n = \frac{1}{2} m v_n^2 = \frac{Z^2 e^4 m}{8 \epsilon_0^2 h^2 n^2} $

where:

• $ m $ is the mass of the electron
• $ Z $ is the atomic number of the nucleus
• $ e $ is the charge of the electron
• $ \epsilon_0 $ is the permittivity of free space
• $ h $ is Planck's constant
• $ n $ is the principle quantum number, which is 2 in this case

For a hydrogen atom (H) ($ Z = 1 $) in the $ n = 2 $ state:

$ KE_{H (n=2)} = \frac{1^2 \cdot e^4 m}{8 \epsilon_0^2 h^2 \cdot 2^2} = \frac{e^4 m}{32 \epsilon_0^2 h^2} $

For a singly ionized helium ion (He$^+$) ($ Z = 2 $) in the $ n = 2 $ state:

$ KE_{He^+ (n=2)} = \frac{2^2 \cdot e^4 m}{8 \epsilon_0^2 h^2 \cdot 2^2} = \frac{4 e^4 m}{32 \epsilon_0^2 h^2} = \frac{e^4 m}{8 \epsilon_0^2 h^2} $

To find the ratio of the kinetic energy of the $ n=2 $ electron for the H atom to that of the He$^+$ ion, we divide the kinetic energies:

$ \text{Ratio} = \frac{KE_{H (n=2)}}{KE_{He^+ (n=2)}} = \frac{\frac{e^4 m}{32 \epsilon_0^2 h^2}}{\frac{e^4 m}{8 \epsilon_0^2 h^2}} = \frac{1}{4} $

Therefore, the ratio of the kinetic energy of the $ n=2 $ electron for the H atom to that of the He$^+$ ion is:

Option A: $\frac{1}{4}$

When binding energy per nucleon increases for a nuclear process, energy is released.

(a) For 1 < A < 50, on fusion, mass number of the resulting nucleus will be less than 100. No energy will be released.

(b) For 51 < A < 100, on fusion, mass number of the resulting nucleus will be between 100 and 200. As BE increases, energy will be released.

(c) On fission for 100 < A < 200, the mass number for fission nuclei will be between 50 and 100. As BE decreases, no energy will be released.

(d) On fission for 200 < A < 260, the mass number for the fission nuclei will be between 100 and 130. Since BE will increase, energy will be released.

Iodine and Yttrium are medium sized nuclei and therefore, have more binding energy per nucleon as compared to uranium which has a big nuclei and less binding energy per nucleon. So that Iodine and Yttrium are more stable and therefore possess less energy and less restmass. When uranium nuclei explodes, it will convert into I and Y nuclei having kinetic energies.

For largest $\lambda$ in UV $\to$ 122 nm

For $\Delta$E$_{\mathrm{min}}$,

${1 \over {{\lambda _{lyman}}}} = {R_c}\left( {1 - {1 \over 4}} \right)$

[$\because$ ${1 \over \lambda } = {R_c}\left( {{1 \over {n_2^2}} - {1 \over {n_1^2}}} \right)$]

${1 \over {{\lambda _{lyman}}}} = {{3{R_c}} \over 4}$ ..... (i)

R$_c$ = Rydber constant = 1.097 $\times$ 10$^7$ m$^{-1}$

For min $\lambda$ in IR region, n = D $\to$ n = 3

${1 \over {{\lambda _{IR}}}} = {{{R_c}} \over 9}$ ...... (ii)

On dividing eq. (i) by (ii)

${1 \over {{\lambda _{lyman}}}} \times {\lambda _{IR}} = {1 \over {{{3{R_c}} \over 4}}} \times {9 \over {{R_c}}} = {{3{R_c}} \over 4} \times {9 \over {{R_c}}} = {{27} \over 4}$

$\therefore$ ${\lambda _{IR}} = {{27} \over 4} \times 122$ nm = 823.5 nm

In a nuclear fusion reaction, two or more lighter nuclei combine to give a comparatively heavier nucleus and some matter is converted into energy.

In a nuclear fission reaction, a heavy nuclear breaks into two or more lighter nuclei and some matter is converted into energy.

$\beta$-decay essentially proceeds by weak nuclear forces and converts some matter into energy. Exothermic nuclear reaction is possible for both type of nuclei (with low and high atomic number) and releases energy.

(A) Using

$R=R_{0} A^{1 / 3}$

$\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$

$\frac{\mathrm{R}}{\mathrm{R}_{\mathrm{He}}}=\left(\frac{\mathrm{A}}{4}\right)^{1 / 3}$

$(14)^{1 / 3}=\left(\frac{\mathrm{A}}{4}\right)^{1 / 3}$

$A=14 \times 4=56$

So, $Z=(56-30)=26$

$=\text { atomic number }$

(B) Frequency of $K_{\alpha}$ line $=f$

$\therefore f=\mathrm{R} c(z-b)^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right], b=1$ for $\mathrm{K} \alpha, z=26$

$f=\left(1.1 \times 10^{7}\right)\left(3 \times 10^{8}\right)(26-1)\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]$

$f=\frac{3.3 \times 10^{15} \times 25 \times 25 \times 3}{4}$

$f=1.547 \times 10^{18} \mathrm{~Hz}$

$\therefore$ Frequency of $\mathrm{K} \alpha$ line of $\mathrm{X}$-rays

$=1.547 \times 10^{18} \mathrm{~Hz}=1.55 \times 10^{18} \mathrm{~Hz} \text {. }$