A radioactive material P first decays into Q and then Q decays to non-radioactive material R. Which of the following figure represents time dependent mass of P, Q and R?

A.
Radioactivity
B.
Radioactivity
C.
Radioactivity
D.
Radioactivity

Solution & Explanation

Related Formula

N = N₀ e-λ t

where $λ$ $\lambda$ is the decay constant.

Core Logic

Initially, only material P is present, so its mass decreases exponentially from a maximum value to zero. Material Q is formed from P and then decays into R, so its mass initially increases from zero, reaches a maximum, and then decreases to zero. Material R is stable and accumulated over time, so its mass increases continuously from zero and levels off at a maximum value equal to the initial mass of P.

Step 1: Graphical Identification

Looking at the options, option (2) correctly depicts the exponential decay of P, the transient rise and fall of Q, and the continuous growth of R to a stable value.

Radioactive decay curves for P, Q, and R

Pattern Recognition

For sequential decay $P arrow Q arrow R$ $P \rightarrow Q \rightarrow R$, parent $P$ $P$ always starts at max and drops to 0. Intermediate $Q$ $Q$ starts at 0, peaks, and returns to 0. Final stable product $R$ $R$ starts at 0 and grows asymptotically to max value.

Chapter Mix

Class 12 Physics: Nuclei

Radioactive decay curves for P, Q, and R

Radioactive decay curves for P, Q, and R

Reference Study Guides

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

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