Question

Two radioactive substances A and B of mass numbers 200 and 212 respectively, shows spontaneous \(\alpha\)-decay with same Q value of 1 MeV. The ratio of energies of \(\alpha\)-rays produced by A and B is ________.

• A. \(\frac{2548}{2650}\)
• B. \(\frac{2706}{2646}\)
• C. \(\frac{2597}{2600}\)
• D. \(\frac{2862}{2499}\)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
During an alpha decay, the total released energy (Q-value) is shared between the emitted alpha particle and the recoiling daughter nucleus. Due to conservation of momentum, the lighter alpha particle carries away the vast majority of the kinetic energy.
Step 2: Key Formula or Approach:
The kinetic energy of the emitted $\alpha$-particle is derived from momentum conservation and is given by:
$E_\alpha = Q \left( \frac{A - 4}{A} \right)$
where $A$ is the mass number of the parent nucleus.
Step 3: Detailed Explanation:
For substance A:
Mass number $A_1 = 200$.
The energy of the $\alpha$-particle from A is:
$E_{\alpha 1} = Q \left( \frac{200 - 4}{200} \right) = Q \left( \frac{196}{200} \right) = Q \left( \frac{49}{50} \right)$.
For substance B:
Mass number $A_2 = 212$.
The energy of the $\alpha$-particle from B is:
$E_{\alpha 2} = Q \left( \frac{212 - 4}{212} \right) = Q \left( \frac{208}{212} \right) = Q \left( \frac{52}{53} \right)$.
The problem asks for the ratio of their energies $\frac{E_{\alpha 1}}{E_{\alpha 2}}$:
$\text{Ratio} = \frac{Q (49 / 50)}{Q (52 / 53)} = \frac{49}{50} \times \frac{53}{52}$.
Multiply the numerators and denominators:
Numerator: $49 \times 53 = 2597$.
Denominator: $50 \times 52 = 2600$.
Therefore, the ratio is $\frac{2597}{2600}$.
Step 4: Final Answer:
The ratio of energies is $\frac{2597}{2600}$.