Question

An alpha particle with kinetic energy K just initiates a nuclear reaction with element X. This being an endothermic reaction with Q-value of magnitude $E_0$. In another nuclear reaction neutron with same kinetic energy K initiates same nuclear reaction, which is endothermic reaction of Q value of magnitude $\frac{19}{16}E_0$. Atomic mass of X is

Answer 1

15 u

Answer 2

The threshold kinetic energy ($K_{th}$) of a projectile particle with mass $m_p$ incident on a target nucleus with mass $m_T$ to initiate an endothermic nuclear reaction with Q-value $Q$ is given by: $K_{th} = -Q \left(1 + \frac{m_p}{m_T}\right)$

In this problem, the target nucleus is element X, with atomic mass $m_X$. The projectile in the first case is an alpha particle ($\alpha$) with mass $m_\alpha$. The reaction is endothermic with Q-value magnitude $E_0$, so $Q_\alpha = -E_0$. The kinetic energy $K$ initiates the reaction, meaning $K$ is the threshold energy for the alpha particle: $K = -Q_\alpha \left(1 + \frac{m_\alpha}{m_X}\right) = E_0 \left(1 + \frac{m_\alpha}{m_X}\right)$ --- (1)

In the second case, the projectile is a neutron ($n$) with mass $m_n$. The reaction is endothermic with Q-value magnitude $\frac{19}{16}E_0$, so $Q_n = -\frac{19}{16}E_0$. The kinetic energy $K$ initiates the same nuclear reaction, implying $K$ is also the threshold energy for the neutron: $K = -Q_n \left(1 + \frac{m_n}{m_X}\right) = \frac{19}{16}E_0 \left(1 + \frac{m_n}{m_X}\right)$ --- (2)

For JEE and NEET exams, the atomic masses of common particles are often approximated as: Mass of alpha particle ($m_\alpha$) $\approx 4$ atomic mass units (u). Mass of neutron ($m_n$) $\approx 1$ atomic mass unit (u).

Equating the expressions for $K$ from (1) and (2): $E_0 \left(1 + \frac{m_\alpha}{m_X}\right) = \frac{19}{16}E_0 \left(1 + \frac{m_n}{m_X}\right)$

Since $E_0 > 0$ (as it's a magnitude of Q-value for an endothermic reaction), we can cancel $E_0$: $1 + \frac{m_\alpha}{m_X} = \frac{19}{16} \left(1 + \frac{m_n}{m_X}\right)$

Substitute the approximate masses $m_\alpha = 4$ u and $m_n = 1$ u: $1 + \frac{4}{m_X} = \frac{19}{16} \left(1 + \frac{1}{m_X}\right)$ $1 + \frac{4}{m_X} = \frac{19}{16} + \frac{19}{16m_X}$

Rearrange the terms to solve for $m_X$: $\frac{4}{m_X} - \frac{19}{16m_X} = \frac{19}{16} - 1$

Combine terms on both sides: $\frac{4 \times 16 - 19}{16m_X} = \frac{19 - 16}{16}$ $\frac{64 - 19}{16m_X} = \frac{3}{16}$ $\frac{45}{16m_X} = \frac{3}{16}$

Multiply both sides by $16m_X$: $45 = 3m_X$

Solve for $m_X$: $m_X = \frac{45}{3} = 15$

Therefore, the atomic mass of element X is 15 atomic mass units.