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:

  1. (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 $Z_1 Z_2$.



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