Atoms - Result Question 2
2. An alpha nucleus of energy $\frac{1}{2} m v^{2}$ bombards a heavy nuclear target of charge $Z e$. Then the distance of closest approach for the alpha nucleus will be proportional to
[2010]
(a) $\frac{1}{Ze}$
(b) $v^{2}$
(c) $\frac{1}{m}$
(d) $\frac{1}{v^{4}}$
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Answer:
Correct Answer: 2. (c)
Solution:
- (c) Kinetic energy of alpha nucleus is equall to electrostatic potential energy of the system of the alpha particle and the heavy nucleus. That is,
$\frac{1}{2} m v^{2}=\frac{1}{4 \pi \varepsilon_0} \frac{q _{\alpha} Z e}{r_0}$
where $r_0$ is the distance of closest approach
$r_0=\frac{2}{4 \pi \varepsilon_0} \frac{q _{\alpha} Z e}{m v^{2}}$
$\Rightarrow r_0 \propto Z e \propto q _{\alpha} \propto \frac{1}{m} \propto \frac{1}{v^{2}}$
Hence, correct option is (c). (a) The kinetic energy of the projectile is given by
$\frac{1}{2} mv^{2}=\frac{Ze(2 e)}{4 \pi \varepsilon_0 r_0}$
$=\frac{Z_1 Z_2}{4 \pi \varepsilon_0 r_0}$
Thus energy of the projectile is directly proportional to $Z_1 Z_2$.
