Question

An alpha particle moves toward a rest nucleus if k.e is 10 MeV and atomic no of nucleus is 50.the closest approach will 

Answer 1

Answer:

The closest approach will be $1.44\times10^{-14}\ m$

Explanation:

Given that,

Kinetic energy K.E = 10 MeV

Charge on nucleus = 50e C

Let the alpha particle be at in infinity charge on alpha particle is 2e.

The kinetic energy will be convert to potential energy at maximum approach

Kinetic energy = potential energy

$\dfrac{1}{2}mv^2=\dfrac{kq_{1}q_{2}}{r}$

$10^{7}eV=\dfrac{k\times2e\times50e}{r}$

$10^{7}eV=\dfrac{k100e^2}{r}\ V$

$1=\dfrac{9\times10^4\times1.6\times10^{-19}}{r}$

$r = 9\times10^{4}\times1.6\times10^{-19}$

$r = 1.44\times10^{-14}\ m$

Hence, The closest approach will be $1.44\times10^{-14}\ m$