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Question: An alpha nucleus of energy \(\dfrac{1}{2} mv^2\) bombards a heavy nuclear target of charge Ze. Then ...

An alpha nucleus of energy 12mv2\dfrac{1}{2} mv^2 bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to
A. v2{v^2}
B. 1m\dfrac{1}{m}
C. 1v4\dfrac{1}{{{v^4}}}
D. 1Ze\dfrac{1}{{Ze}}

Explanation

Solution

The sum of kinetic and potential energies of an alpha particle is equal to the sum of kinetic and potential energies of the nuclear target.

Complete step by step answer:
An alpha nucleus of kinetic energy (K1)\left( {{K_1}} \right) is 12mv2\dfrac{1}{2}m{v^2} bombards a heavy nuclear target of charge q2=Ze{q_2} = Ze where Z represents the atomic number of the element and e is charge of an electron. The atomic number of alpha particles is 2 i.e., atomic number of helium and charge of an alpha particle q1=2e{q_1} = 2e.
Initially, the target is at rest. So, the kinetic energy (K2)\left( {{K_2}} \right) of the target is zero and when the alpha particle reaches nearer to the nuclear target i.e., the distance of closest approach (d). In this case, the potential energy (P1)\left( {{P_1}} \right) of the alpha particle is zero. The potential energy of the target when an alpha particle approaches the target is P2{P_2}.
According to the conservation of energy, energy can neither be created nor destroyed i.e., sum of initial potential and kinetic energies is equal to sum of final potential and kinetic energies.
P1+K1=P2+K2{P_1} + {K_1} = {P_2} + {K_2}
⇒0+12mv2=q1q24π∈0d+0\Rightarrow 0 + \dfrac{1}{2}m{v^2} = \dfrac{{{q_1}{q_2}}}{{4\pi { \in _0}d}} + 0
⇒12mv2=q1q24π∈0d\Rightarrow \dfrac{1}{2}m{v^2} = \dfrac{{{q_1}{q_2}}}{{4\pi { \in _0}d}}
⇒d=2×2e×Zemv24π∈0\Rightarrow d = \dfrac{{2 \times 2e \times Ze}}{{m{v^2}4\pi { \in _0}}}
⇒d=Ze2mv2π∈0\Rightarrow d = \dfrac{{Z{e^2}}}{{m{v^2}\pi { \in _0}}}
⇒d=K(1m)\Rightarrow d = K\left( {\dfrac{1}{m}} \right) [K=Ze2v2π∈0]\left[ {K = \dfrac{{Z{e^2}}}{{{v^2}\pi { \in _0}}}} \right]
Hence, d is proportional to 1m\dfrac{1}{m} as all the other quantities are constant i.e., K is a constant.

Therefore, option B is correct.

Note: The conservation of energy states that the sum of initial energies is equal to the sum of final energies. The value of d increases with decrease in the mass of the alpha particle and vice-versa.