The mass of a nucleus Z^A X is less than the sum of the masses of (A - Z) number of neutrons and Z n

Q: Why must the mass defect be positive for a spontaneous reaction?

According to Einstein's mass-energy equivalence ($E = mc^2$), a positive mass defect means that some mass has been converted into energy. This release of energy allows the system to transition to a lower, more stable energy state, which is a requirement for a spontaneous process.

Q: Why does the alpha particle get most of the kinetic energy?

Because the parent nucleus is initially at rest, the alpha particle and the daughter nucleus must recoil with equal and opposite momenta to conserve linear momentum. Since kinetic energy is $p^2 / 2m$, the particle with the smaller mass (the alpha particle) will have a much larger kinetic energy.

Q: How did we know the mass of Polonium-210?

In the actual JEE Advanced exam, the mass of Polonium-210 was provided in the table. It was inadvertently missed in some reproductions of the paper, but its mass is $209.982876 \text{ u}$.

Q: Why is proton emission from Polonium-210 not possible?

If you calculate the mass defect for the emission of a proton from Polonium-210 (forming Bismuth-209), you will find that the total mass of the products is greater than the mass of Polonium-210. This means the reaction requires an input of energy and cannot happen spontaneously.

Q: What does 'complete fusion' mean in the context of the options?

Complete fusion means the two reacting nuclei merge entirely to form a single, heavier nucleus without emitting any other particles. We simply add their masses and compare the sum to the mass of the resulting nucleus.

Question

Passage:
The mass of a nucleus $_{Z}^{A}	ext{X}$ is less than the sum of the masses of $(A - Z)$ number of neutrons and $Z$ number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass $M$ can break into two light nuclei of masses $m_1$ and $m_2$ only if $(m_1 + m_2) < M$. Also two light nuclei of masses $m_3$ and $m_4$ can undergo complete fusion and form a heavy nucleus of mass $M'$ only if $(m_3 + m_4) > M'$. The masses of some neutral atoms are given in the table below:

$\begin{array}{llll}
_{1}^{1}	ext{H} & 1.007825	ext{u} & _{1}^{2}	ext{H} & 2.014102	ext{u} \
_{3}^{6}	ext{Li} & 6.01513	ext{u} & _{3}^{7}	ext{Li} & 7.016004	ext{u} \
_{64}^{152}	ext{Gd} & 151.919803	ext{u} & _{82}^{206}	ext{Pb} & 205.974455	ext{u} \
_{1}^{3}	ext{H} & 3.016050	ext{u} & _{2}^{4}	ext{He} & 4.002603	ext{u} \
_{30}^{70}	ext{Zn} & 69.925325	ext{u} & _{34}^{82}	ext{Se} & 81.916709	ext{u} \
_{84}^{210}	ext{Po} & 209.982876	ext{u} & & 
\end{array}$

Question 1:
The correct statement is
(A) The nucleus $_{3}^{6}	ext{Li}$ can emit an alpha particle.
(B) The nucleus $_{84}^{210}	ext{Po}$ can emit a proton.
(C) Deuteron and alpha particle can undergo complete fusion.
(D) The nuclei $_{30}^{70}	ext{Zn}$ and $_{34}^{82}	ext{Se}$ can undergo complete fusion.

Question 2:
The kinetic energy (in keV) of the alpha particle, when the nucleus $_{84}^{210}	ext{Po}$ at rest undergoes alpha decay, is
(A) 5316
(B) 5422
(C) 5707
(D) 5818

Sigma Percentile · Accepted Answer

### The Core Principle: Mass Defect and Energy

To understand whether a nuclear reaction can occur spontaneously, we must look at the fundamental principle of mass-energy equivalence, famously encapsulated in Einstein's equation, $E = mc^2$. The universe inherently favors states of lower energy. For a nuclear reaction (be it fission or fusion) to proceed without external energy input, the total mass of the reacting nuclei must be strictly greater than the total mass of the resulting products.

This difference in mass is called the **mass defect** ($\Delta m$). When $\Delta m$ is positive, the 'lost' mass is converted into kinetic energy, which is released into the surroundings. If $\Delta m$ is negative, the reaction would require an input of energy to occur, making it non-spontaneous.

### Analyzing the First Question: Spontaneous Reactions

Let's systematically evaluate the options provided in the first question by calculating their mass defects.

**Option (a):** Can Lithium-6 emit an alpha particle?
If $_{3}^{6}	ext{Li}$ were to emit an alpha particle ($_{2}^{4}	ext{He}$), it would leave behind a Deuteron ($_{1}^{2}	ext{H}$).
$$\Delta m = M_{	ext{Li}} - (M_{	ext{He}} + M_{	ext{H}^2})$$
$$\Delta m = 6.01513 - (4.002603 + 2.014102) = -0.001575 	ext{ u}$$
Since the mass defect is negative, this decay is kinematically forbidden.

**Option (c):** Can a Deuteron and an alpha particle undergo complete fusion?
This is the exact reverse of the reaction in option (a). 
$$_{1}^{2}	ext{H} + _{2}^{4}	ext{He} ightarrow _{3}^{6}	ext{Li}$$
$$\Delta m = (M_{	ext{H}^2} + M_{	ext{He}}) - M_{	ext{Li}}$$
$$\Delta m = (2.014102 + 4.002603) - 6.01513 = +0.001575 	ext{ u}$$
Because the mass defect is positive, energy is released, making this fusion reaction perfectly possible. Thus, **Option (c) is the correct statement**.

*(For completeness, checking option (d) reveals that fusing Zinc-70 and Selenium-82 to form Gadolinium-152 also results in a negative mass defect, meaning it cannot happen spontaneously.)*

### The Second Question: Alpha Decay of Polonium-210

Now, let's tackle the second question, which asks for the kinetic energy of the alpha particle emitted during the decay of Polonium-210.

The decay equation is:
$$_{84}^{210}	ext{Po} ightarrow _{82}^{206}	ext{Pb} + _{2}^{4}	ext{He}$$

First, we calculate the mass defect:
$$\Delta m = M_{	ext{Po}} - (M_{	ext{Pb}} + M_{	ext{He}})$$
$$\Delta m = 209.982876 - (205.974455 + 4.002603) = 0.005818 	ext{ u}$$

This mass defect is converted into the total energy released, known as the **Q-value**:
$$Q = \Delta m 	imes 931.5 	ext{ MeV/u}$$
$$Q = 0.005818 	imes 931.5 \approx 5.419 	ext{ MeV} = 5419 	ext{ keV}$$

### The Final Calculation: Sharing the Energy

This total energy ($5419 	ext{ keV}$) is shared between the alpha particle and the recoiling Lead nucleus as kinetic energy. Since the Polonium nucleus was initially at rest, the law of conservation of linear momentum dictates that the alpha particle and the Lead nucleus must fly apart with equal and opposite momenta ($p_{\alpha} = p_{	ext{Pb}}$).

Kinetic energy $K$ is related to momentum $p$ by the equation $K = \frac{p^2}{2m}$. Because their momenta are equal, the kinetic energy is inversely proportional to the mass ($K \propto \frac{1}{m}$). This means the lighter alpha particle will carry away the vast majority of the energy.

We can use the energy sharing formula to find the exact kinetic energy of the alpha particle:
$$K_{\alpha} = \left(\frac{m_{	ext{Pb}}}{m_{	ext{Pb}} + m_{\alpha}}ight) Q$$
$$K_{\alpha} = \left(\frac{206}{206 + 4}ight) 	imes 5419 	ext{ keV}$$
$$K_{\alpha} = \left(\frac{206}{210}ight) 	imes 5419 	ext{ keV} \approx 5316 	ext{ keV}$$

Therefore, the alpha particle shoots out with a kinetic energy of **$5316 	ext{ keV}$**, making **Option (a)** the correct answer.

Animated Solution for Physics - Atoms and Nuclei: Comprehension Passage

The correct statement is

Select Answer:

The kinetic energy (in keV) of the alpha particle, when the nucleus $_{84}^{210} Po$ at rest undergoes alpha decay, is

Select Answer:

Visualized Solution

\text{Analyzing the Passage}

\text{Evaluating Option (a)}

\text{Evaluating Option (c)}

\text{Evaluating Option (d)}

\text{Alpha Decay of Polonium-210}

\text{Kinetic Energy of Alpha Particle}

The Sigma Insight: Nucleus and Nuclear Reaction

The Core Principle

Analyzing the First Question

The Second Question

The Final Calculation

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Animated Solution for Physics - Atoms and Nuclei: Comprehension Passage

The correct statement is

Select Answer:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The kinetic energy (in keV) of the alpha particle, when the nucleus 84210​Po at rest undergoes alpha decay, is

Select Answer:

Visualized Solution

\text{Analyzing the Passage}

\text{Evaluating Option (a)}

\text{Evaluating Option (c)}

\text{Evaluating Option (d)}

\text{Alpha Decay of Polonium-210}

\text{Kinetic Energy of Alpha Particle}

The Sigma Insight: Nucleus and Nuclear Reaction

The Core Principle

Analyzing the First Question

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The Final Calculation

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