Question

An alpha particle having charge 10Mev An alpha-particle having energy 10Mev
collides with a nucleus of
Atomic number 50 . Then distance of
​

Answer 1

Answer:

The closest approach will be 1.44\times10^{-14}\ m1.44×10

−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}

2

1

mv

2

=

r

kq

1

q

2

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

7

eV=

r

k×2e×50e

10^{7}eV=\dfrac{k100e^2}{r}\ V10

7

eV=

r

k100e

2

V

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

r

9×10

4

×1.6×10

−19

r = 9\times10^{4}\times1.6\times10^{-19}r=9×10

4

×1.6×10

−19

r = 1.44\times10^{-14}\ mr=1.44×10

−14

m

Hence, The closest approach will be 1.44\times10^{-14}\ m1.44×10

−14

m