An atom 83X is bombarded by shower of fundamental particles and in 10 s this atom absorbed 10 electrons, 10 protons and…

Solution & Explanation

Related Formula

$R = R₀ A1/3$ $R = R_0 A^{1/3}$ Surface area $S \propto R² \propto A2/3$ $S \propto R^2 \propto A^{2/3}$

Core Logic

Nucleons consist of protons and neutrons. Calculate the initial and final total nucleon number $A$ $A$ to determine the ratio of surface areas. Note: Electrons are not nucleons and do not contribute to the nuclear surface area.

Step 1: Initial State

Initial number of nucleons $Aᵢ = 8$ $A_i = 8$. Initial surface area $Sᵢ \propto (Aᵢ)2/3 = 82/3 = 4K$ (where K is a constant of proportionality).

Step 2: Final State

The atom absorbs 10 protons and 9 neutrons. Final number of nucleons $Af = 8 + 10 + 9 = 27$ $A_f = 8 + 10 + 9 = 27$. Final surface area $Sf \propto (Af)2/3 = 272/3 = 9K$ .

Step 3: Percentage Growth

Percentage increase in surface area:

% increase = (Sf - Sᵢ)/(Sᵢ) \times 100

= (9K - 4K)/(4K) \times 100

= (5)/(4) \times 100 = 125%

Step 4: Final Conclusion

The calculated percentage growth is $125%$ $125\%$, which is not in the options. Our Ans. (BONUS) NTA Ans. (3) which is $225%$ $225\%$ (This would imply $(9K)/(4K) = 2.25$ $\frac{9K}{4K} = 2.25$ or $225%$ $225\%$ as the ratio $Sf/Sᵢ$ $S_f/S_i$, not the growth).

Pattern Recognition

Radius scales as $A1/3$ $A^{1/3}$, Volume scales as $A$ $A$, Surface Area scales as $A2/3$ $A^{2/3}$. Watch out for wording like 'growth' vs 'final ratio' which exam bodies sometimes confuse.

Chapter Mix

Class 12 Physics: Nuclei

More Nuclei Previous-Year Questions — Page 5

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

An atom ⁸₃X${}^{8}_{3}X$ is bombarded by shower of fundamental particles and in 10 s this atom absorbed 10 electrons, 10 protons and 9 neutrons. The percentage growth in the surface area of the nucleons is recorded by:

Solution & Explanation

Related Formula

Core Logic

Step 1: Initial State

Step 2: Final State

Step 3: Percentage Growth

Step 4: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Nuclei Previous-Year Questions — Page 5

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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An atom $⁸₃X$ ${}^{8}_{3}X$ is bombarded by shower of fundamental particles and in 10 s this atom absorbed 10 electrons, 10 protons and 9 neutrons. The percentage growth in the surface area of the nucleons is recorded by: