Choose the correct nuclear process from the below options [p: proton, n: neutron, e- : electron, e+ : positron, v…

Q: Choose the correct nuclear process from the below options [p: proton, n: neutron, e- : electron, e+ : positron, v: neutrino, nu : antineutrino]

In basic beta^- emission processes, a neutron decays inside a nucleus to satisfy lepton numbers and conservation rules: Charge Balance: 0 arrow (+1) + (-1) + 0 = 0 (Conserved) Lepton Family Index: 0 arrow 0 + (+1) + (-1) = 0 (Conserved via antineutrino entry). This perfectly isolates option (1). Negative beta emission is always accompanied by an antineutrino, whereas positive positron transformation releases a regular neutrino molecule. Class 12 Physics: Nuclei

Q49 jee_main_2024_29_jan_morning Nuclear Fusion and Binding Energy

The explosive in a Hydrogen bomb is a mixture of

₁H²

,

₁H³

and

₃Li⁶

in some condensed form. The chain reaction is given by:

arrayl _ 3 L i ^ 6 + _ 0 n ^ 1 arrow _ 2 H e ^ 4 + _ 1 H ^ 3 _ 1 H ^ 2 + _ 1 H ^ 3 arrow _ 2 H e ^ 4 + _ 0 n ^ 1 array

During the explosion the energy released is approximately: [Given:

M(Li) = 6.01690 ~amu

,

M(₁H²) = 2.01471 ~amu

,

M(₂He⁴) = 4.00388 ~amu

, and

1 ~amu = 931.5 ~MeV

]

A. 28.12 MeV
B. 12.64 MeV
C. 16.48 MeV
D. 22.22 MeV

Solution

Related Formula

The Q-value or energy released ( $Q$ $Q$) during a nuclear reaction sequence is determined from mass defect ( $Δ m$ $\Delta m$):

Q = Δ m \times 931.5 ~MeV

Core Logic

Adding the two equations together to obtain the single combined net nuclear reaction:

₃Li⁶ + ₀n¹ + ₁H² + ₁H³ arrow 2(₂He⁴) + ₁H³ + ₀n¹

Cancelling intermediate species appearing on both sides yields:

₃Li⁶ + ₁H² arrow 2(₂He⁴)

Step 1: Calculate Mass Defect

The mass defect $Δ m$ $\Delta m$ of this net process is:

Δ m = M(Li) + M(₁H²) - 2 M(₂He⁴)

Substituting the given mass profiles:

Δ m = 6.01690 + 2.01471 - 2(4.00388)

Δ m = 8.03161 - 8.00776 = 0.02385 ~amu

Step 2: Compute Energy Released

Converting mass defect into MeV value:

Q = 0.02385 \times 931.5 ~MeV \approx 22.216 ~MeV

Rounding off gives approximately $22.22 ~MeV$ $22.22 \mathrm{~MeV}$.

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.

Chapter Mix

Class 12 Physics: Nuclei

Q35 jee_main_2024_30_january_evening Nuclear Fission and Mass Defect

In a nuclear fission reaction of an isotope of mass

M

, three similar daughter nuclei of same mass are formed. The speed of a daughter nuclei in terms of mass defect

Δ M

will be :

A. $\sqrt((2 c Δ M)/(M))$ $\sqrt{\frac{2 c \Delta M}{M}}$
B. $Δ M c²3$ $\frac{\Delta \mathrm{M} c^{2}}{3}$
C. $c \sqrt((2 Δ M)/(M))$ $c \sqrt{\frac{2 \Delta M}{M}}$
D. $c \sqrt((3 Δ M)/(M))$ $c \sqrt{\frac{3 \Delta M}{M}}$

Solution

Related Formula

Q = Δ M c²

Q = Σ K.E.products

Core Logic

The nuclear fission reaction can be written as:

(X) arrow (Y) + (Z) + (P)

The parent mass is $M$ $M$. Three similar daughter nuclei are formed, each with mass $\approx (M)/(3)$ $\approx \frac{M}{3}$. The total energy released due to the mass defect $Δ M$ $\Delta M$ is $Δ M c²$ $\Delta M c^2$. This energy is equally distributed among the three identical daughter nuclei as kinetic energy (assuming parent is at rest).

Step 1: Equate Energy

Δ M c² = (1)/(2) ((M)/(3)) V² + (1)/(2) ((M)/(3)) V² + (1)/(2) ((M)/(3)) V²

Δ M c² = 3 \times (1)/(2) ((M)/(3)) V²

Δ M c² = (1)/(2) M V²

Step 2: Solve for V

V² = (2 Δ M c²)/(M)

V = c \sqrt((2 Δ M)/(M))

Pattern Recognition

Since total mass of the products is $M$ $M$ (ignoring the tiny mass defect for kinetic energy calculations), the total kinetic energy $(1)/(2) M V²$ $\frac{1}{2} M V^2$ equals the released energy $Δ M c²$ $\Delta M c^2$. The number of identical fragments doesn't change the velocity expression.

Chapter Mix

Class 12 Physics: Nuclei

Q45 jee_main_2024_31_jan_evening Nuclear Radius

The mass number of nucleus having radius equal to half of the radius of nucleus with mass number

192

is:

A. $24$ $24$
B. $32$ $32$
C. $40$ $40$
D. $20$ $20$

Solution

Related Formula

Nuclear radius is empirically related to mass number by: $R = R₀ A1/3$ $R = R_0 A^{1/3}$

Core Logic

Given $R₁ = (R₂)/(2)$ $R_1 = \frac{R_2}{2}$ where $A₂ = 192$ $A_2 = 192$. We need to find $A₁$ $A_1$.

Step 1: Forming the Ratio

(R₁)/(R₂) = ((A₁)/(A₂))1/3

(1)/(2) = ((A₁)/(192))1/3

Step 2: Cubing Both Sides

((1)/(2))³ = (A₁)/(192)

(1)/(8) = (A₁)/(192)

A₁ = (192)/(8) = 24

Pattern Recognition

Since $R \propto A1/3$ $R \propto A^{1/3}$, scaling $R$ $R$ by $k$ $k$ means scaling $A$ $A$ by $k³$ $k^3$. Half the radius ( $k = 1/2$ $k = 1/2$) means $1/8$ $1/8$th the mass number.

Chapter Mix

Class 12 Physics: Nuclei

Q60 jee_main_2024_31_jan_evening Nuclear Size and Density

A nucleus has mass number

A₁

and volume

V₁

. Another nucleus has mass number

A₂

and volume

V₂

. If relation between mass number is

A₂ = 4A₁

, then

(V₂)/(V₁) =

________.

Numerical Answer. Answer: 4 to 4

Solution

Related Formula

$R = R₀ A1/3$ $R = R_0 A^{1/3}$

V = (4)/(3)π R³

Core Logic

Since radius $R$ $R$ is proportional to $A1/3$ $A^{1/3}$, the volume $V$ $V$ (which depends on $R³$ $R^3$) will be directly proportional to the mass number $A$ $A$.

Step 1: Show Proportionality

V = (4)/(3)π (R₀ A1/3)³ = (4)/(3)π R₀³ A

This proves that $V \propto A$ $V \propto A$.

Step 2: Calculate Ratio

(V₂)/(V₁) = (A₂)/(A₁)

Given that $A₂ = 4A₁$ $A_2 = 4A_1$:

(V₂)/(V₁) = (4A₁)/(A₁) = 4

Pattern Recognition

Nuclear density is constant for all nuclei. Therefore, Mass $\propto$ $\propto$ Volume. Since Mass number ( $A$ $A$) represents mass, Volume is strictly linearly proportional to Mass number.

Chapter Mix

Class 12 Physics: Nuclei

Q60 jee_main_2024_31_jan_morning Mass Defect And Energy

The mass defect in a particular reaction is

0.4 g

. The amount of energy liberated is

n \times 10⁷ kWh

where

n =

_______. (speed of light

= 3 \times 10⁸ m/s

)

Numerical Answer. Answer: 1 to 1

Solution

Related Formula

E = Δ m c²

1 kWh = 3.6 \times 10⁶ J

Core Logic

Given the mass defect:

Δ m = 0.4 g = 0.4 \times 10⁻³ kg

The total energy liberated in Joules is:

E = (0.4 \times 10⁻³) \times (3 \times 10⁸)²

E = 0.4 \times 10⁻³ \times 9 \times 10¹⁶

E = 3.6 \times 10¹³ J

Step 2: Conversion to kWh

We need the answer in $kWh$ $\mathrm{kWh}$. Since $1 kWh = 1000 W \times 3600 s = 3.6 \times 10⁶ J$ :

E = 3.6 \times 10¹³3.6 \times 10⁶ kWh

E = 10⁷ kWh

Comparing this to $n \times 10⁷ kWh$ $n \times 10^7\mathrm{\,kWh}$, we get: $n = 1$ $n = 1$

Chapter Mix

Class 12 Physics: Nuclei

Choose the correct nuclear process from the below options [p: proton, n: neutron, e⁻$\mathbf{e}^{-}$ : electron, e⁺$\mathrm{e}^{+}$ : positron, v: neutrino, ν$\overline{\nu}$ : antineutrino]

Solution & Explanation

Core Logic

Step 1: Conservation Cross-Check

Pattern Recognition

Chapter Mix

Reference Study Guides

More Nuclei Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Calculate Mass Defect

Step 2: Compute Energy Released

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Equate Energy

Step 2: Solve for V

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Forming the Ratio

Step 2: Cubing Both Sides

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Show Proportionality

Step 2: Calculate Ratio

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Conversion to kWh

Chapter Mix

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Choose the correct nuclear process from the below options [p: proton, n: neutron, $e⁻$ $\mathbf{e}^{-}$ : electron, $e⁺$ $\mathrm{e}^{+}$ : positron, v: neutrino, $ν$ $\overline{\nu}$ : antineutrino]