Question

An α particle of energy 5Mev is scattered through 180° by a stationary uranium nucleus. The distance of closest approach is of the order of:

Answer 1

$\huge\underline\blue{\sf Answer:}$

$\red{\boxed{\sf Closest\:Approach=5.3×{10}^{-12}cm}}$

$\huge\underline\blue{\sf Solution:}$

$\large\underline\pink{\sf Given: }$

• Energy (E ) = 5MeV

• $\sf{{e}^{-}=1.6×{10}^{-19}C}$

$\large\underline\pink{\sf To\:Find: }$

• Closest approch (r) = ?

━━━━━━━━━━━━━━━━━━━━━━━━━

At closest approch all the Kinetic Energy (KE) of the $\alpha$ -particle will be converted into Potential Energy (PE) of the system

i.e. KE = PE

$\large{♡}$$\large{\boxed{\sf {\frac{1}{2}}m{v}^{2}={\frac{1}{4πE_o}}{\frac{q_1q_2}{r}}}}$

Here ,

$\sf{{\frac{1}{4πE_o}}=9×{10}^{9}}$

$\large{\sf{\frac{1}{2}}m{v}^{2}=5Mev }$

$\large{\sf {e}^{-}=1.6×{10}^{-19}C}$

On putting value :

$\large\implies{\sf 5Mev={\frac{9×{10}^{9}×2e×92e}{r}}}$

$\large\implies{\sf r={\frac{9×{10}^{9}×2×92×(1.6×{10}^{-19})}{5×{10}^{6}×1.6×{10}^{-19}}}}$

$\large\implies{\sf r= 5.3×{10}^{-14}m }$

Or

$\large\implies{\sf r=5.3×{10}^{-12}cm}$

$\huge\red{♡}$$\red{\boxed{\sf Closest\:Approach=5.3×{10}^{-12}cm}}$