Question

An α particle with energy of 3 MeV is moving towards a nucleus of ⁵⁰Sn. Its minimum distance of approach to the nucleus is 𝑓×10⁻¹⁴ m. The value of 𝑓 is ________ (rounded off to one decimal place).

Answer 1

Correct Answer: 4.7 - 4.9

Given:
Energy of α–particle: \( E = 3 \text{ MeV} \) 
Target nucleus: \( ^{50}\text{Sn} \) (Z = 50)
Charge of α–particle: \( Z_{\alpha} = 2 \)

Formula (Distance of closest approach):
\[ d = \frac{1}{4\pi \epsilon_0}\,\frac{Z_{\alpha} Z e^2}{E} \] Using, \[ \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \quad,\quad e = 1.6 \times 10^{-19} \text{ C} \] Convert energy to joule: \[ E = 3 \text{ MeV} = 3 \times 10^6 \times 1.6\times10^{-19} = 4.8\times10^{-13} \text{ J} \] Substitute values: \[ d = 9 \times 10^9 \times \frac{(2)(50)(1.6\times10^{-19})^2}{4.8\times10^{-13}} \] Simplify numerator: \[ (2)(50)=100 \] \[ (1.6\times10^{-19})^2 = 2.56\times10^{-38} \] \[ 100 \times 2.56\times10^{-38} = 2.56\times10^{-36} \] Now: \[ d = 9\times10^9 \times \frac{2.56\times10^{-36}}{4.8\times10^{-13}} \] Divide: \[ \frac{2.56}{4.8} = 0.5333 \] \[ \frac{10^{-36}}{10^{-13}} = 10^{-23} \] Thus, \[ d = 9 \times 10^9 \times 0.5333\times10^{-23} \] Multiply: \[ 9 \times 0.5333 = 4.7997 \] \[ d = 4.8 \times 10^{-14} \text{ m} \] Therefore:
\[ f = 4.8 \] Final Answer: \( f \approx 4.7\text{ to }4.9 \)