Question

The distance of closest approach of an alpha particle fired at the nucleus with momentum P is d. The distance of closest approach when the alpha particle is fired at same nucleus with momentum 3P will be:
A. $3d$
B. $\dfrac{d}{3}$
C. $9d$
D. $\dfrac{d}{9}$

Answer 1

Hint- For solving this numerical we will firstly know the alpha particles and what happens to it when it reaches the nucleus. Alpha rays or alpha particles are the positively charged particles. A highly energetic helium nucleus which contains two protons and two neutrons is called the alpha-particle. The penetration power of the Alpha particles is low but the ionization power of the alpha particles is high.

Step-By-Step answer:
When alpha particles reaches to nucleus:
When the alpha particle reaches the closest distance of approach, its total kinetic energy will be converted into the potential energy of the system.
Thus ${E_i} = \dfrac{{{P^2}}}{{2m}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{Ze \times q}}{r}$
Where r is the distance of closest approach and q is the charge of the incident particle.
Thus for particles with different momenta, $r\alpha \dfrac{1}{{{P^2}}}$
Thus for $\alpha$ particle with momentum 3P, the distance of the closest approach will be
Where ${r_1} = d,{r_2} = ?,{P_1} = P,{P_2} = 3P$
Using the proportionality relation given above we get
$r\alpha \dfrac{1}{{{P^2}}} \\\ \Rightarrow \dfrac{{{r_1}}}{{{r_2}}} = \dfrac{{{P_2}^2}}{{{P_1}^2}} \\\ \Rightarrow \dfrac{d}{{{r_2}}} = \dfrac{{{{\left( {3P} \right)}^2}}}{{{P^2}}} \\\ \Rightarrow {r_2} = \dfrac{d}{9} \\\$
Hence the correct option is “D”.

Note- The alpha particle is a fast-moving packet carrying two protons and two neutrons (a helium nucleus). Alpha particles bear a charge of +2 and interact intensely with matter. In general, alpha particles have a very limited ability to penetrate other materials. In other words, these particles of ionizing radiation can be blocked by a sheet of paper, skin, or even a few inches of air.