Atoms and Nuclei

To determine the minimum number of days after which the laboratory can be considered safe for use, we need to understand how the radioactive material decays over time.

The half-life of a radioactive material is the time it takes for half of the material to decay. For our given material, the half-life is 18 days. This means every 18 days, the amount of radioactive material is reduced to half of its previous amount.

Given that the initial radiation level is 64 times more than the permissible level, we can represent this situation mathematically. Let $ R $ be the initial radiation level and $ P $ be the permissible level of radiation. We are given:

$ R = 64P $

After each half-life period of 18 days, the amount of radioactive material will be halved. If $ n $ is the number of half-life periods required for the radiation to reduce to the permissible level, we can write:

$ \frac{R}{2^n} = P $

Using the given information ($ R = 64P $), we get:

$ \frac{64P}{2^n} = P $

Dividing both sides by $ P $, we get:

$ \frac{64}{2^n} = 1 $

This implies:

$ 2^n = 64 $

Knowing that $ 64 $ is a power of $ 2 $:

$ 64 = 2^6 $

Therefore, $ n = 6 $. This means it will take six half-life periods for the radiation to decay to a safe level.

Since each half-life period is 18 days, the total number of days required is:

$ 6 \times 18 = 108 $

Hence, the minimum number of days after which the laboratory can be considered safe for use is 108 days.

Answer: Option C (108)

The binding energies of $_7^{15}N$ and $_8^{15}O$ are given by

$B{E_N} = \Delta {m_N}{c^2} = (8{m_n} + 7{m_p} - {M_N}){c^2}$

$ = (8 \times 1.008665 + 7 \times 1.007825 - 15.000109 \times 931.5$

= 115.49 MeV,

$B{E_O} = \Delta {m_O}{c^2} = (7{m_n} + 8{m_p} - {M_O}){c^2}$

$ = (7 \times 1.008665 + 8 \times 1.007825 - 15.003065 \times 931.5$

= 111.95 MeV.

The difference in binding energies of $_7^{15}N$ and $_8^{15}O$ is

$\Delta BE = B{E_N} - B{E_O} = 115.49 - 111.95$

= 3.54 MeV. ........ (1)

The electrostatic energies of $_7^{15}N$ and $_8^{15}O$ are given by

${E_N} = {3 \over 5}{{Z(Z - 1){e^2}} \over {4\pi { \in _0}R}} = {3 \over 5}{{7(7 - 1)1.44} \over R}$

$ = {{36.288} \over R}$ MeV-fm

${E_O} = {3 \over 5}{{8(8 - 1)1.44} \over R} = {{48.384} \over R}$ MeV-fm.

The difference in electrostatic energies of $_7^{15}N$ and $_8^{15}O$ is

$\Delta E = {E_O} - {E_N} = (12.096/R)$ ...... (2)

Since, $\Delta E = \Delta BE$, equations (1) and (2) give $R = 12.096/3.54 = 3.42$ fm.

The nuclear fusion is responsible for energy production in stars via fusion of hydrogen nuclei into helium nuclei. In sun, the fusion takes place dominantly by proton-proton cycle, $4_1^1H \to _2^4He + 2{e^ + } + 2\nu + 2\gamma $. The neutrino ($\nu $) and $\gamma$-rays emissions are parts of this fusion reaction.

The uranium based fission reactions involve absorption of thermal neutrons by ${}_{92}^{235}$U nuclei to produce the highly fissionable ${}_{92}^{236}$U nuclei. This nuclei then fissions into two parts e.g., ${}_{92}^{236}$U $\to$ ${}_{63}^{137}$I + ${}_{39}^{97}$Y + 2n. The fission fragments are unstable and undergo $\beta$-decay to reduce their neutron to proton ratio. The fragments are generally formed in excited states and consequently emit $\gamma$-rays. The heavy water (D₂O) is used as a moderator to slow down the fast moving neutrons.

In $\beta$-decay, a neutron is converted into a proton. In this process, an electron and an antineutrino are created and emitted from the nucleus, n $\to$ p + e^$-$ + $\overline v $. The $\beta$-decay in ${}_{27}^{60}$Co $\to$ ${}_{28}^{60}$Ni + e^$-$ + $\overline v $. The daughter nuclei ${}_{28}^{60}$Ni is formed in excited state and comes to ground state by $\gamma$-ray emission.

The $\gamma$-rays are high energy electromagnetic rays. These rays are generally emitted when a nuclei in excited state (high energy) makes a transition to a lower state (low energy).

${}_{92}^{236}U \to {}_{54}^{140}Xe + {}_{38}^{94}Sr + x + y$

K_x = 2 MeV, K_y = 2 MeV, K_Xe = ?, K_Sr = ?

By conservation of charge number and mass number, x $\equiv$ y $\equiv$ n

B.E. per nucleon of ${}_{92}^{236}U = 7.5$ MeV

B.E. per nucleon of ${}_{54}^{140}Xe$ or ${}_{38}^{94}Sr = 8.5$ MeV

Q value of reaction,

Q = Net kinetic energy gained in the process

$ = {K_{Xe}} + {K_{Sr}} + 2 + 2 - 0 = {K_{Xe}} + {K_{Sr}} + 4$ ..... (1)

As number of nucleons is conserved in a reaction, so Q = Difference of binding energies of the nuclei

$ = 140 \times 8.5 + 94 \times 8.5 - 236 \times 7.5 = 219$ MeV ...... (2)

From eqns. (i) and (ii)

${K_{Xe}} + {K_{Sr}} = 219 - 4 = 215$ MeV ....... (3)

The linear momentum of a particle of mass m and kinetic energy K is given by $p = \sqrt {2mK} $. Since the masses and kinetic energies of x and y are very small in comparison to that of ${}_{54}^{140}Xe$ and ${}_{38}^{94}Sr$, we can neglect the linear momentum of these particles. Initially, the linear momentum of ${}_{92}^{236}U$ is zero (at rest). Finally, the products ${}_{54}^{140}Xe$ and ${}_{38}^{94}Sr$ will move in opposite direction with equal linear momentum (by conservation of linear momentum). Thus,

$\sqrt {2{M_{Xe}}{K_{Xe}}} = \sqrt {2{M_{Sr}}{K_{Sr}}} $, i.e.,

$140{K_{Xe}} = 94{K_{Sr}}$. ...... (4)

Solve equations (3) and (4) to get K_Xe = 86 MeV and K_Sr = 129 MeV.

Let’s carefully match them:

• (a) Davisson and Germer Experiment → Demonstrated wave nature of electrons via electron diffraction. → (i)

• (b) Millikan’s Oil Drop Experiment → Measured charge of an electron. → (ii)

• (c) Rutherford Experiment → Showed the existence of nucleus in the atom through alpha-particle scattering. → (iv)

• (d) Franck–Hertz Experiment → Demonstrated quantisation of energy levels in atoms (mercury). → (iii)

So correct matching is:

(a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)

✅ Answer: Option B

• Fresh wood: 20 decays/min (current activity if alive / just cut).

• Old museum sample: 2 decays/min.

• Half-life of C‑14 = 5730 years.

• Find the age of wooden piece in museum.

Step 1: Ratio of activities

$ \frac{A}{A_0} = \frac{2}{20} = 0.1 $

So, the museum sample’s activity is 10% of that in fresh wood.

Step 2: Radioactive decay law

$ A = A_0 e^{-\lambda t} $

$ \frac{A}{A_0} = e^{-\lambda t} $

So,

$ 0.1 = e^{-\lambda t} $

$ t = \frac{\ln(0.1)}{-\lambda} $

Step 3: Decay constant

$ \lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{5730 \,\text{yr}} \approx 1.2097 \times 10^{-4} \,\text{yr}^{-1} $

Step 4: Solve for $t$

$ t = \frac{\ln(0.1)}{-1.2097 \times 10^{-4}} $

$ \ln(0.1) = -2.3026 $

$ t = \frac{2.3026}{1.2097 \times 10^{-4}} $

$ t \approx 19039 \,\text{years} $

✅ Final Answer:

Option C: 19039 years

The wavelength of the K_$\alpha$ X-ray line is related to the atomic number of the element by Moseley's law, which can be expressed as:

$ \sqrt{\nu} = A(Z - B) $

Here, $\nu$ is the frequency of the emitted X-ray, $Z$ is the atomic number, and $A$ and $B$ are constants. Given that the wavelength $\lambda$ is inversely proportional to the frequency $\nu$ (since $\nu = \frac{c}{\lambda}$, where $c$ is the speed of light), we can rewrite this equation in terms of wavelength:

$ \lambda \propto \frac{1}{(Z - B)^2} $

Now, let's consider the ratio of the wavelengths for copper (Cu) and molybdenum (Mo):

$ \frac{\lambda_{Cu}}{\lambda_{Mo}} = \left(\frac{Z_{Mo} - B}{Z_{Cu} - B}\right)^2 $

For the K_$\alpha$ X-ray line, the constant $B$ can be approximated as 1. Using the atomic numbers $Z_{Cu} = 29$ and $Z_{Mo} = 42$, we can plug these values into the ratio:

$ \frac{\lambda_{Cu}}{\lambda_{Mo}} = \left(\frac{42 - 1}{29 - 1}\right)^2 = \left(\frac{41}{28}\right)^2 $

Calculating the above expression:

$ \frac{41}{28} \approx 1.464 $

$ \left(1.464\right)^2 \approx 2.14 $

Therefore, the ratio $\frac{\lambda_{Cu}}{\lambda_{Mo}}$ is close to 2.14. The correct option is:

Option B 2.14

We have

$m(_1^2H) + m(_2^4He) = 2.014102 + 4.002603 = 6.016705\,u$

$m(_3^6Li) = 6.015123\,u$

${m_1} + {m_2} > M$

Thus, option (A) is wrong.

$m(_1^1H) + m(_{83}^{209}Bi) = 1.007825\,u + 208.980388\,u = 209.988213\,u$

$m(_{84}^{210}Po) = 209.982876\,u$

${m_1} + {m_2} > M$

Thus, option (B) is wrong.

$m(_1^2H) + m(_2^4He) = 2.014102\,u + 4.002603\,u = 6.016705\,u$

$_3^6Li = 6.015123\,u$

$({m_3} + {m_4}) > M'$

Thus, option (C) is correct. Therefore, deuteron and alpha particle can go complete fusion.

$m(_{30}^{70}Zn) + _{34}^{82}Se = 69.925325\,u + 81.916709\,u = 151.842034\,u$

$_{64}^{152}Gd = 151.919803\,u$

${m_3} + {m_4} < M'$

Thus, option (D) is wrong.

The alpha decay of $_{84}^{210}Po$ is given by

$_{84}^{210}Po \to _{82}^{206}Pb + _2^4He$

The energy released during this process is

$Q = ({M_{Po}} - {M_{Pb}} - {M_{He}}){c^2}$

$ = (209.982876 - 205.974455 - 4.002603)\,u \times {c^2}$

$ = (0.005818\,u){c^2} = (0.005818\,u) \times 932\,MeV$

$ = 5.422\,MeV = 5422\,keV$

Kinetic energy of a particle, ${K_\alpha } = {{(A - 4)Q} \over A}$

${K_\alpha } = {{(210 - 4)} \over {210}} \times 5422\,keV = {{206} \over {210}} \times 5422\,keV = 5319\,keV$

In alpha decay, atomic number decreases by 2 and mass number decreases by 4.

P $\to$ 2

In $\beta$⁺ decay, atomic number decreases by 1 and mass number remains unchanged.

Q $\to$ 1

In fission, heavy nucleus breaks into two light nuclei.

R $\to$ 4

In proton emission, both atomic number and mass number decreases by 1.

S $\to$ 3

The orbital angular momentum is

$L = {{nh} \over {2\pi }}$

${{3h} \over {2\pi }} = {{nh} \over {2\pi }} \Rightarrow n = 3$

The radius of the orbit is

$4.5({a_0}) = {a_0}\left( {{{{n^2}} \over Z}} \right)$

$Z = {{{n^2}} \over {4.5}} = {{{3^2}} \over {4.5}} = {9 \over {4.5}} = 2$

Thus, the possible transitions are 3$\to$2, 3$\to$1 and 2$\to$1. For the transition 3$\to$2, the wavelength is

${1 \over \lambda } = R{(2)^2}\left[ {{1 \over 4} - {1 \over 9}} \right] = 4R\left[ {{{9 - 4} \over {36}}} \right] = {{5R} \over 9}$

$ \Rightarrow \lambda = {9 \over {5R}}$

For the transition 3$\to$1, the wavelength is

${1 \over \lambda } = R{(2)^2}\left[ {1 - {1 \over 9}} \right] = 4R\left( {{8 \over 9}} \right) = {{32R} \over 9}$

$ \Rightarrow \lambda = {9 \over {32R}}$

For the transition 2$\to$1, the wavelength is

${1 \over \lambda } = R{(2)^2}\left[ {1 - {1 \over 4}} \right] = 4R\left( {{3 \over 4}} \right) = 3R$

$ \Rightarrow \lambda = {1 \over {3R}}$

${K_p} + {K_{\overline e }} + {K_{\overline v }}$ = 0.8 $\times$ 10⁶ eV

When electron has zero kinetic energy is shared by antineutrino and proton.

Then, ${K_p} + {K_{\overline v }}$ = 0.8 $\times$ 10⁶ eV

As antineutrino is very light mass in comparison to proton so it will have almost contribution in total energy.

$\therefore$ Its energy is almost 0.8 $\times$ 10⁶ eV

Total energy remains conserved. Energy is shared by antineutrino, proton and electron. Kinetic energy of electron has continuous spectrum and it is maximum when antineutrino does not share any kinetic energy.

So total energy is shared with proton and electron only.

$\therefore$ K $\le$ 0.8 $\times$ 10⁶ eV

and kinetic energy of electron will be minimum or zero when total energy is shared by proton and antineutrino.

$\therefore$ 0 $\le$ K $\le$ 0.8 $\times$ 10⁶ eV

${1 \over \lambda } = {R_H}{Z^2}\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]$

For singly-ionized helium atom, Z = 2. For hydrogen atom Z = 1.

For Balmer series n₁ = 2.

For first spectral line of hydrogen :

${1 \over {6561}} = {R_H} \times {(1)^2}\left[ {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right] = {{5R} \over {36}}$

$ \Rightarrow {R_H} = {{36} \over {5 \times 6561}}$

For second spectral line of helium,

${1 \over \lambda } = {R_H} \times {(2)^2}\left[ {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right] = {{3{R_H}} \over 4}$

$ = {3 \over 4} \times {{36} \over {5 \times 6561}}$

$ \Rightarrow \lambda = 1215\,\mathop A\limits^o $

To find the quantized rotational energy of a diatomic molecule, we can use Bohr's quantization condition for angular momentum. Bohr's quantization condition states that the angular momentum $ L $ of an electron in a hydrogen atom is quantized and given by:

$ L = n \hbar $

where $ n $ is a positive integer (known as the quantum number) and $ \hbar $ is the reduced Planck's constant, given by $ \hbar = {\hbar \over {2\pi}} $.

For a diatomic molecule, we extend this concept to its rotational motion. A rigid diatomic molecule can be approximated as a rigid rotor with a moment of inertia $ I $. The angular momentum $ L $ for the molecule in the nth level is given by:

$ L = n \hbar $

where $ n $ is the quantum number (note that $ n = 0 $ is not allowed).

The rotational kinetic energy for a rotating body with moment of inertia $ I $ is given by:

$ E = {1 \over 2} I \omega^2 $

where $ \omega $ is the angular velocity. The angular momentum $ L $ is related to the angular velocity $ \omega $ by:

$ L = I \omega $

Solving for $ \omega $, we get:

$ \omega = {L \over I} $

Substituting this back into the expression for energy, we get:

$ E = {1 \over 2} I \left( {L \over I} \right)^2 = {L^2 \over 2I} $

Using Bohr's quantization condition $ L = n \hbar $, the energy expression becomes:

$ E = {{(n \hbar)^2} \over {2I}} $

Substituting $ \hbar = {\hbar \over {2\pi}} $, we obtain:

$ E = {n^2 \over {2I}} \left( {{\hbar^2} \over {4\pi^2}} \right) = n^2 \left( {{\hbar^2} \over {8\pi^2 I}} \right) $

Therefore, the quantized rotational energy of the diatomic molecule in the nth level is:

$ E_n = n^2 \left( {{\hbar^2} \over {8\pi^2 I}} \right) $

So, the correct option is:

Option D

$ n^2 \left( {{{h^2} \over {8\pi^2 I}}} \right) $

The energy of photon is equal to the energy difference between the ground level and first excited level.

$hv = {E_2} - {E_1}$

$hv = {{(4 - 1){h^2}} \over {8{\pi ^2}I}} \Rightarrow I = {{3h} \over {8{\pi ^2}v}}$

$I = {{3 \times 2\pi \times {{10}^{ - 34}}} \over {[(8{\pi ^2}4)/\pi ] \times {{10}^{11}}}} = {3 \over {16}} \times {10^{ - 45}}$ kg-m² = 1.87 $\times$ 10^$-$46 kg-m²

The moment of inertia of CO molecule is

I = $\mu$r² ....... (i)

where, $\mu$ = reduced mass of the CO molecule, r = distance between C and O or bond length

The reduced mass $\mu$ of the CO molecule is

$\mu = {{{m_1}{m_2}} \over {{m_1} + {m_2}}} = \left[ {{{(12)(16)} \over {12 + 16}}} \right] \times {5 \over 3} \times {10^{ - 27}}$ kg

But $I = 1.87 \times {10^{ - 46}}$ kg m² (From the above question)

From equation (i), we get

${r^2} = {I \over \mu }$

Substituting the values of I and $\mu$ in above equation, we get

${r^2} = {{1.87 \times {{10}^{ - 46}}} \over {\left[ {{{12 \times 16} \over {28}} \times {5 \over 3} \times {{10}^{ - 27}}} \right]}}$

or ${r^2} = {{1.87 \times {{10}^{ - 46}} \times 28 \times 3} \over {12 \times 16 \times 5 \times {{10}^{ - 27}}}}$ or $r = 1.3 \times {10^{ - 10}}$ m

To find the relationship between the speed of the particle and the quantum number $n$, we need to follow a few steps using the given information and fundamental quantum mechanical principles.

Firstly, for a particle moving in a one-dimensional box of length $a$, the allowed wavelengths $\lambda$ of standing waves can be given by:

$ \lambda = \frac{2a}{n} $

for $ n = 1, 2, 3, \ldots $.

The de Broglie relation links the wavelength $\lambda$ of a particle to its momentum $p$:

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

From this, we can express the momentum $p$ in terms of the quantum number $n$:

$ \frac{h}{p} = \frac{2a}{n} $

Solving for $p$, we get:

$ p = \frac{nh}{2a} $

The kinetic energy $E$ of the particle is related to its momentum $p$ by the equation:

$ E = \frac{p^2}{2m} $

Substituting $ p = \frac{nh}{2a} $ into the energy expression, we get:

$ E = \frac{(nh/2a)^2}{2m} = \frac{n^2 h^2}{8ma^2} $

The kinetic energy can also be written in terms of the speed $v$ of the particle:

$ E = \frac{1}{2} m v^2 $

Equating the two expressions for energy, we have:

$ \frac{1}{2} m v^2 = \frac{n^2 h^2}{8ma^2} $

Solving for the speed $v$, we get:

$ v^2 = \frac{n^2 h^2}{4m^2 a^2} $

Therefore, the speed $v$ of the particle is:

$ v = \frac{nh}{2ma} $

From the above expression, we see that the speed $v$ of the particle is directly proportional to $n$:

$ v \propto n $

Therefore, the speed of the particle is proportional to:

Option D: $ n $

In a nuclear fusion reactor, the gas becomes plasma primarily because of the high temperature maintained inside the reactor core. At such high temperatures, typically in the range of millions of degrees Kelvin, the thermal energy is sufficient to ionize the atoms, meaning electrons are stripped from the nuclei, resulting in a collection of positively charged deuteron nuclei and negatively charged electrons. This ionized state of matter is known as plasma.

The strong nuclear force and Coulomb force (between deuterons and between deuteron-electron pairs) play roles in the fusion process and the behavior of particles. However, the ionization into plasma is primarily due to the extremely high temperatures.

Therefore, the correct option is:

Option D: the high temperature maintained inside the reactor core.

To determine the minimum temperature required for two deuteron nuclei to reach a separation of $4 \times 10^{-15}$ m, we need to equate the initial kinetic energy of the deuterons with the potential energy at the specified separation distance.

The initial kinetic energy of each deuteron is given as :

$ KE = 1.5 \, k \, T $

Since there are two deuterons, the total kinetic energy is :

$ KE_{\text{total}} = 2 \times 1.5 \, k \, T = 3 \, k \, T $

The Coulomb potential energy (PE) between two deuterons at a separation distance $r = 4 \times 10^{-15}$ m is given by :

$ PE = \frac{e^2}{4 \pi \epsilon_0 r} $

Using the given value:

$ \frac{e^2}{4 \pi \epsilon_0} = 1.44 \times 10^9 \, \text{eV} \cdot \text{m} $

So, the potential energy becomes :

$ PE = \frac{1.44 \times 10^9 \, \text{eV} \cdot \text{m}}{4 \times 10^{-15} \, \text{m}} $

Simplifying :

$ PE = \frac{1.44 \times 10^9}{4 \times 10^{-15}} \, \text{eV} $

$ PE = 0.36 \times 10^{24} \, \text{eV} $

Now, we equate the total kinetic energy to the potential energy to find the temperature $ T $ :

$ 3 \, k \, T = 0.36 \times 10^{24} \, \text{eV} $

Solving for $ T $ :

$ T = \frac{0.36 \times 10^{24}}{3 \times 8.6 \times 10^{-5}} \, \text{K} $

$ T = \frac{0.36 \times 10^{24}}{2.58 \times 10^{-4}} \, \text{K} $

$ T = 1.395 \times 10^{9} \, \text{K} $

This temperature is in the range :

Option A: $1.0 \times {10^9}K < T < 2.0 < {10^9}K$

So, the correct answer is :

Option A : $1.0 \times {10^9}K < T < 2.0 < {10^9}K$