Atoms and Nuclei

Kinetic energy of ejected electron $=$ Energy of incident photon - Energy required to ionize the electron from $n$th orbit (all in eV )

$ \begin{array}{ll} \therefore & 10.4=\frac{1242}{90}-\left|E_n\right| \\ & 10.4=13.8-\frac{13.6}{n^2} \\ \Rightarrow & \frac{13.6}{n^2}=3.4 \\ & \left(\text { as } E_n \propto \frac{1}{n^2} \text { and } E_1=-13.6 \mathrm{eV}\right) \end{array} $

Solving this equation, we get

               $ n=2 $

The nucleus of ${ }_2 \mathrm{He}^4$ contains 2 neutrons and 2 protons.

Mass of 2 protons $=2 \times 1.0073=2.0146 u$

Mass of 2 neutrons $=2 \times 1.0087=2.0174 u$

Total mass of 2 protons and 2 neutrons

$ =(2.0146+2.0174)=4.032 u $

Mass of helium nucleus $=4.0015 u$

Thus, mass defect is lacking of mass in forming the helium nucleus from 2 protons and 2 neutrons.

$ \begin{aligned} \therefore \Delta m & =\text { Mass defect } \\ & =(4.032-4.0015) \mathrm{u}=0.0305 \mathrm{u} \\ & =0.0305 \mathrm{amu} \end{aligned} $

but $1 \mathrm{amu}=931 \mathrm{MeV}$

Hence, binding energy,

$ \begin{aligned} \Delta E & =(\Delta m) \times 931 \\ & =0.0305 \times 931=28.4 \mathrm{MeV} \end{aligned} $

To determine which transitions of a $\mathrm{He}^{+}$ ion can produce a spectral line with the same wavelength as a given transition in a hydrogen atom, let's examine the Rydberg formula:

For a hydrogen atom ($\mathrm{Z}=1$), the wavelength $\lambda_1$ for a transition from $n_2$ to $n_1$ is given by:

$ \frac{1}{\lambda_1} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) $

Here, $R$ is the Rydberg constant, and $n_1$ and $n_2$ are the principal quantum numbers with $n_2 > n_1$.

For a $\mathrm{He}^{+}$ ion (with $Z=2$), if the electron transitions from $n_4$ to $n_3$, the wavelength $\lambda_2$ is given by:

$ \frac{1}{\lambda_2} = R \cdot 4 \left(\frac{1}{n_3^2} - \frac{1}{n_4^2}\right) $

Assuming $\lambda_1 = \lambda_2$, we have:

$ R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = R \cdot 4 \left(\frac{1}{n_3^2} - \frac{1}{n_4^2}\right) $

By comparing both sides, we get the relationships:

$ n_1 = \frac{n_3}{2}, \quad n_2 = \frac{n_4}{2} $

Both $n_1$, $n_2$, $n_3$, and $n_4$ must be integers, as they represent principal quantum numbers.

This condition is satisfied when $n_3 = 2$ and $n_4 = 4$, ensuring that transitions corresponding to these quantum numbers yield the same wavelength for both $\mathrm{He}^{+}$ and hydrogen.

Potential energy of an electron

$(U)=-2 \text { (kinetic Energy K) }$

So, total energy, $E=U+K=-2 K+K$

$\begin{aligned} E & =-K \\ \Rightarrow K & =-E \Rightarrow K=-(-1.51 \mathrm{~eV})=1.51 \mathrm{~eV} \end{aligned}$

To determine the half-life of the radioactive isotope, we need to use the information given and the formula for radioactive decay. The formula for radioactive decay is:

$N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{\frac{1}{2}}}}$

where:

• $N(t)$ is the remaining quantity of the isotope at time $t$.
• $N_0$ is the initial quantity of the isotope.
• $T_{\frac{1}{2}}$ is the half-life of the isotope.
• $t$ is the elapsed time.

According to the given information, after 2 hours (which is 120 minutes), one-eighth of the starting amount of the isotope remains undecayed. Therefore, if $\frac{N(t)}{N_0} = \frac{1}{8}$, we can write:

$\frac{N(t)}{N_0} = \left( \frac{1}{2} \right)^{\frac{t}{T_{\frac{1}{2}}}}$

Substituting the values we have:

$\frac{1}{8} = \left( \frac{1}{2} \right)^{\frac{120}{T_{\frac{1}{2}}}}$

We know that $\frac{1}{8}$ can be written as $\left( \frac{1}{2} \right)^3$, therefore:

$\left( \frac{1}{2} \right)^3 = \left( \frac{1}{2} \right)^{\frac{120}{T_{\frac{1}{2}}}}$

By equating the exponents, we get:

$3 = \frac{120}{T_{\frac{1}{2}}}$

Solving for $T_{\frac{1}{2}}$:

$T_{\frac{1}{2}} = \frac{120}{3} = 40 \ \text{minutes}$

Thus, the half-life of the isotope is 40 minutes. Therefore, the correct option is:

Option B: 40 min

From radioactive decay law,

$ - {{dN} \over {dt}} \propto N$ or $ - {{dN} \over {dt}} = \lambda N$

Thus, $A = - {{dN} \over {dt}}$ or $A = \lambda N$

$ \Rightarrow A = \lambda {N_0}{e^{ - \lambda t}}$ ...... (i)

Where, ${A_0} = \lambda {N_0}$ is the activity of the radioactive material at time, t = 0

At time, ${A_1} = {A_0}\,{e^{ - \lambda {t_1}}}$ ..... (ii)

At time, ${A_2} = {A_0}\,{e^{ - \lambda {t_2}}}$ ..... (iii)

On dividing Eq. (ii) by Eq. (iii), we have

${{{A_1}} \over {{A_2}}} = {{{e^{ - \lambda {t_1}}}} \over {{e^{ - \lambda {t_2}}}}} = {e^{ - \lambda ({t_1} - {t_2})}}$

$ \Rightarrow {A_1} = {A_2}\,{e^{ - \lambda ({t_1} - {t_2})}}$

Kinetic energy of a proton is given by

$E = {1 \over 2}{m_p}v_p^2$

$ \Rightarrow E = {{{p^2}} \over {2{m_p}}}$ [$\because$ $p = mv$]

$\therefore$ Wavelength of proton is, ${\lambda _1} = {h \over p}$

$ \Rightarrow {\lambda _1} = {h \over {\sqrt {2{m_p}E} }}$ ..... (i)

Now, energy of photon is given by

$E = h{v_2} \Rightarrow E = {{hc} \over {{\lambda _2}}}$

$ \Rightarrow {\lambda _2} = {{hc} \over E}$ ....... (ii)

where, $\lambda$₂ is wavelength of photon.

On dividing Eq. (ii) by Eq. (i), we get

${{{\lambda _2}} \over {{\lambda _1}}} = {{hc} \over E} \times {{\sqrt {2{m_p}E} } \over h}$

$ \Rightarrow {{{\lambda _2}} \over {{\lambda _1}}} = {c \over {\sqrt E }}\sqrt {2{m_p}} $

$ \Rightarrow {{{\lambda _2}} \over {{\lambda _1}}} \propto {E^{ - 1/2}}$

Given, radius of hydrogen atom, r = 2.5 $\times$ 10^$-$13 m

Volume of one atom, $V = {4 \over 3}\pi {r^3}$

Number of atoms in 1 mole = 6.023 $\times$ 10²³

Volume of 1 mole of H-atoms = $N \times {4 \over 3}\pi {r^3}$

$ = 6.023 \times {10^{23}} \times {4 \over 3} \times 3.14 \times {(2.5 \times {10^{ - 13}})^3}$

$ = 3.94 \times {10^{ - 14}}$ m³

According to law of radioactivity, the rate of disintegration is given as

$R = \lambda N = \lambda {N_0}{e^{ - \lambda t}}$

Given, $R = 5000$

$ \Rightarrow 5000 = \lambda {N_0}{e^{ - \lambda t}}$ ...... (i)

After 5 min, $R = 1250$

$\therefore$ $1250 = \lambda {N_0}{e^{ - \lambda (t + 5)}}$ ...... (ii)

On dividing Eq. (i) by Eq. (ii), we get

${{5000} \over {1250}} = {{{e^{ - \lambda t}}} \over {{e^{ - \lambda (t + 5)}}}}$

$ \Rightarrow 4 = {e^{5\lambda }}$

Taking log both sides, we get

${\log _e}4 = {\log _e}({e^{5\lambda }})$

$ \Rightarrow lo{g_e}4 = 5\lambda \Rightarrow \lambda = {1 \over 5}{\log _e}{2^2}$

$ \Rightarrow \lambda = 0.4{\log _e}2$

For hydrogen-like atom,

${1 \over {\lambda '}} = R{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$ ..... (i)

Similarly, for hydrogen atom,

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

For the element to have hydrogen-like spectrum, the ratio of the wavelengths for transition of electron between any two levels n₁ and n₂ should be constant.

Dividing eq. (ii) by eq. (i), we get

${{\lambda '} \over \lambda } = {{R\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)} \over {R{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)}}$ or ${{\lambda '} \over \lambda } = {1 \over {{Z^2}}}$

It is given that wavelength of the hydrogen atom is four times the wavelength of the hydrogen-like atom, i.e.

$\lambda = 4\lambda '$

Hence, $Z = 2$

Now, the element with atomic number 2 is helium.

Number of active nuclei after time t in a sample of radioactive substance is given by

N = N₀e^{$-$$\lambda$ t}

where, N₀ = initial number of nuclei at t = 0 and $\lambda$ = decay constant.

Here, at t = 0, number of nuclei in sample X and Y are equal,

i.e., (N₀)_X = (N₀)_Y = N₀

Also, $\lambda$_X = 4$\lambda$ and $\lambda$_Y = $\lambda$

So, after time t, number of active nuclei in X and Y are

${N_X} = {N_0}{e^{ - 4\lambda t}}$ and ${N_Y} = {N_0}{e^{ - 2\lambda t}}$

$\therefore$ ${{{N_X}} \over {{N_Y}}} = {e^{ - 2\lambda t}}$

${1 \over {{e^3}}} = {1 \over {{e^{2\lambda t}}}}$ [Given, ${{{N_X}} \over {{N_Y}}} = {1 \over {{e^3}}}$]

$ \Rightarrow 2\lambda t = 3 \Rightarrow t = {3 \over {2\lambda }}$