Atoms and Nuclei Previous Year Questions — JEE Main Physics
23 past-year Atoms and Nuclei questions from JEE Main (Physics) — page 3.
Q49 (2024)
The explosive in a Hydrogen bomb is a mixture of ${}_{1}\mathrm{H}^{2}$, ${}_{1}\mathrm{H}^{3}$ and ${}_{3}\mathrm{Li}^{6}$ in some condensed form. The chain reaction is given by:
$$\begin{array}{l} { } _ { 3 } \mathrm { L i } ^ { 6 } + { } _ { 0 } \mathrm { n } ^ { 1 } \rightarrow { } _ { 2 } \mathrm { H e } ^ { 4 } + { } _ { 1 } \mathrm { H } ^ { 3 } \\ { } _ { 1 } \mathrm { H } ^ { 2 } + { } _ { 1 } \mathrm { H } ^ { 3 } \rightarrow { } _ { 2 } \mathrm { H e } ^ { 4 } + { } _ { 0 } \mathrm { n } ^ { 1 } \end{array}$$
During the explosion the energy released is approximately:
[Given: $\mathbf{M}(\mathrm{Li}) = 6.01690 \mathrm{~amu}$, $\mathbf{M}({}_1\mathbf{H}^2) = 2.01471 \mathrm{~amu}$, $\mathbf{M}({}_2\mathbf{H}e^4) = 4.00388 \mathrm{~amu}$, and $1 \mathrm{~amu} = 931.5 \mathrm{~MeV}$]
- 28.12 MeV
- 12.64 MeV
- 16.48 MeV
- 22.22 MeV
### Related Formula
The Q-value or energy released ($Q$) during a nuclear reaction sequence is determined from mass defect ($\Delta m$):
$$Q = \Delta m \times 931.5 \mathrm{~MeV}$$
### Core Logic
Adding the two equations together to obtain the single combined net nuclear reaction:
$${}_{3}\mathrm{Li}^{6} + {}_{0}\mathrm{n}^{1} + {}_{1}\mathrm{H}^{2} + {}_{1}\mathrm{H}^{3} \rightarrow 2\left({}_{2}\mathrm{He}^{4}\right) + {}_{1}\mathrm{H}^{3} + {}_{0}\mathrm{n}^{1}$$
Cancelling intermediate species appearing on both sides yields:
$${}_{3}\mathrm{Li}^{6} + {}_{1}\mathrm{H}^{2} \rightarrow 2\left({}_{2}\mathrm{He}^{4}\right)$$
### Step 1: Calculate Mass Defect
The mass defect $\Delta m$ of this net process is:
$$\Delta m = M(\mathrm{Li}) + M({}_{1}\mathrm{H}^{2}) - 2 M({}_{2}\mathrm{He}^{4})$$
Substituting the given mass profiles:
$$\Delta m = 6.01690 + 2.01471 - 2(4.00388)$$
$$\Delta m = 8.03161 - 8.00776 = 0.02385 \mathrm{~amu}$$
### Step 2: Compute Energy Released
Converting mass defect into MeV value:
$$Q = 0.02385 \times 931.5 \mathrm{~MeV} \approx 22.216 \mathrm{~MeV}$$
Rounding off gives approximately $22.22 \mathrm{~MeV}$.
### Pattern Recognition
When chain equations share intermediate steps (like neutron consumption/generation or tritium tracking), add the algebraic steps together to deduce the overall net target process. This cuts out unnecessary individual constituent mass balances.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Nuclei
Q35 (2024)
In a nuclear fission reaction of an isotope of mass $\mathrm{M}$, three similar daughter nuclei of same mass are formed. The speed of a daughter nuclei in terms of mass defect $\Delta \mathrm{M}$ will be :
- $\sqrt{\frac{2 c \Delta M}{M}}$
- $\frac{\Delta \mathrm{M} c^{2}}{3}$
- $c \sqrt{\frac{2 \Delta M}{M}}$
- $c \sqrt{\frac{3 \Delta M}{M}}$
### Related Formula
$$Q = \Delta M c^2$$
$$Q = \sum K.E._{\text{products}}$$
### Core Logic
The nuclear fission reaction can be written as:
$$(\mathrm{X}) \rightarrow (\mathrm{Y}) + (\mathrm{Z}) + (\mathrm{P})$$
The parent mass is $M$. Three similar daughter nuclei are formed, each with mass $\approx \frac{M}{3}$.
The total energy released due to the mass defect $\Delta M$ is $\Delta M c^2$. This energy is equally distributed among the three identical daughter nuclei as kinetic energy (assuming parent is at rest).
### Step 1: Equate Energy
$$\Delta M c^2 = \frac{1}{2} \left(\frac{M}{3}\right) V^2 + \frac{1}{2} \left(\frac{M}{3}\right) V^2 + \frac{1}{2} \left(\frac{M}{3}\right) V^2$$
$$\Delta M c^2 = 3 \times \frac{1}{2} \left(\frac{M}{3}\right) V^2$$
$$\Delta M c^2 = \frac{1}{2} M V^2$$
### Step 2: Solve for V
$$V^2 = \frac{2 \Delta M c^2}{M}$$
$$V = c \sqrt{\frac{2 \Delta M}{M}}$$
### Pattern Recognition
Since total mass of the products is $M$ (ignoring the tiny mass defect for kinetic energy calculations), the total kinetic energy $\frac{1}{2} M V^2$ equals the released energy $\Delta M c^2$. The number of identical fragments doesn't change the velocity expression.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Nuclei
Q45 (2024)
The mass number of nucleus having radius equal to half of the radius of nucleus with mass number $192$ is:
- $24$
- $32$
- $40$
- $20$
### Related Formula
Nuclear radius is empirically related to mass number by:
$R = R_0 A^{1/3}$
### Core Logic
Given $R_1 = \frac{R_2}{2}$ where $A_2 = 192$.
We need to find $A_1$.
### Step 1: Forming the Ratio
$$\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3}$$
$$\frac{1}{2} = \left(\frac{A_1}{192}\right)^{1/3}$$
### Step 2: Cubing Both Sides
$$\left(\frac{1}{2}\right)^3 = \frac{A_1}{192}$$
$$\frac{1}{8} = \frac{A_1}{192}$$
$$A_1 = \frac{192}{8} = 24$$
### Pattern Recognition
Since $R \propto A^{1/3}$, scaling $R$ by $k$ means scaling $A$ by $k^3$. Half the radius ($k = 1/2$) means $1/8$th the mass number.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Nuclei
Q60 (2024)
A nucleus has mass number $A_1$ and volume $V_1$. Another nucleus has mass number $A_2$ and volume $V_2$. If relation between mass number is $A_2 = 4A_1$, then $\frac{V_2}{V_1} =$ ________.
### Related Formula
$R = R_0 A^{1/3}$
$$V = \frac{4}{3}\pi R^3$$
### Core Logic
Since radius $R$ is proportional to $A^{1/3}$, the volume $V$ (which depends on $R^3$) will be directly proportional to the mass number $A$.
### Step 1: Show Proportionality
$$V = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A$$
This proves that $V \propto A$.
### Step 2: Calculate Ratio
$$\frac{V_2}{V_1} = \frac{A_2}{A_1}$$
Given that $A_2 = 4A_1$:
$$\frac{V_2}{V_1} = \frac{4A_1}{A_1} = 4$$
### Pattern Recognition
Nuclear density is constant for all nuclei. Therefore, Mass $\propto$ Volume. Since Mass number ($A$) represents mass, Volume is strictly linearly proportional to Mass number.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Nuclei
Q60 (2024)
The mass defect in a particular reaction is $0.4\mathrm{\ g}$. The amount of energy liberated is $n \times 10^7\mathrm{\ kWh}$ where $n =$ _______. (speed of light $= 3 \times 10^8\mathrm{\ m/s}$)
### Related Formula
$$E = \Delta m c^2$$
$$1 \text{ kWh} = 3.6 \times 10^6 \text{ J}$$
### Core Logic
Given the mass defect:
$$\Delta m = 0.4\mathrm{\,g} = 0.4 \times 10^{-3}\mathrm{\,kg}$$
The total energy liberated in Joules is:
$$E = (0.4 \times 10^{-3}) \times (3 \times 10^8)^2$$
$$E = 0.4 \times 10^{-3} \times 9 \times 10^{16}$$
$$E = 3.6 \times 10^{13}\mathrm{\,J}$$
### Step 2: Conversion to kWh
We need the answer in $\mathrm{kWh}$.
Since $1\mathrm{\,kWh} = 1000\mathrm{\,W} \times 3600\mathrm{\,s} = 3.6 \times 10^6\mathrm{\,J}$:
$$E = \frac{3.6 \times 10^{13}}{3.6 \times 10^6}\mathrm{\,kWh}$$
$$E = 10^7\mathrm{\,kWh}$$
Comparing this to $n \times 10^7\mathrm{\,kWh}$, we get:
$n = 1$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Nuclei
