Question

An `alpha` - particle of energy 5 MeV is scattered through 180° by gold nucleus. The distance of closest approach is of the order of ______.

विकल्प

• 10^-14 cm

• 10^-10 cm

• 10^-16 cm

• 10^-12 cm

Answer 1

An `alpha` - particle of energy 5 MeV is scattered through 180° by gold nucleus. The distance of closest approach is of the order of 10^-12 cm.

Explanation:

`"r"_0 = 1/ (4pi omega_0) (2"Ze"^2)/"E"`

E = 5 MeV `= 5 xx 10^6 xx 1.6  xx 10^-19` J,

For gold, Z = 79

`"r"_0 = (9 xx 10^9 xx 2 xxx 79 xx (1.6 xx 10^-19)^2)/(5 xx 10^6 xx 1.6  xx 10^-19`

= `4.55 xx 10 ^-14` m

= `4.55 xx 10 ^-12` cm