Question

The maximum K.E found in alpha particle of natural origin is $7.7$ MeV. Find the closest distance of approach for the gold nucleus $(Z=79$ for gold$)$

[Note : At closest distance of approach, kinetic energy is converted into potential energy]

• A. $3\times {10}^{-12}$ m
• B. $3\times {10}^{-13}$ m
• C. $3\times {10}^{-14}$ m
• D. $3\times {10}^{-16}$ m

Answer 1

The correct option is A $3\times {10}^{-12}$ m

$K.E_{max}=P.E$

$7.7\times {10}^6\times 1.6\times {10}^{-19}=Z\frac{q^2}{r^2}$

$12.32\times {10}^{-13}$ =$\frac{79\times (1.6\times {10}^{-19}{)}^2}{r^2}$

$r^2=16.415\times {10}^{-25}$

$r=3\times {10}^{-12}$m

Therefore, the closest distance of approach is $3\times {10}^{-12}m$