When an α-particle of mass 'm' moving with velocity 'v' bombards on a heavy nucleus of charge 'Ze', its distance of closest approach from the nucleus depends on m as :

Question

When an α-particle of mass 'm' moving with velocity 'v' bombards on a heavy nucleus of charge 'Ze', its distance of closest approach from the nucleus depends on m as : (A) (1/√m) (B) (1/m2) (C) m (D) (1/m)

Tardigrade Admin · Accepted Answer

At closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy. ⇒ (1/2) mv2 = (KQq/d) ⇒ d propto (1/m)

Q. When an $α$ -particle of mass 'm' moving with velocity 'v' bombards on a heavy nucleus of charge 'Ze', its distance of closest approach from the nucleus depends on m as :

$\frac{1}{m}$

$\frac{1}{m ^{2}}$

m

$\frac{1}{m}$

At closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy.
$\Rightarrow \frac{1}{2} m v^{2} = \frac{K Qq}{d} \Rightarrow d \propto \frac{1}{m}$

Q. When an α-particle of mass 'm' moving with velocity 'v' bombards on a heavy nucleus of charge 'Ze', its distance of closest approach from the nucleus depends on m as :

m​1​

m21​

m

m1​

At closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy. ⇒21​mv2=dKQq​⇒d∝m1​

Q. When an $α$ -particle of mass 'm' moving with velocity 'v' bombards on a heavy nucleus of charge 'Ze', its distance of closest approach from the nucleus depends on m as :

$\frac{1}{m}$

$\frac{1}{m ^{2}}$

$\frac{1}{m}$

At closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy.
$\Rightarrow \frac{1}{2} m v^{2} = \frac{K Qq}{d} \Rightarrow d \propto \frac{1}{m}$