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GUJCET · Physics · Atoms
An \(\alpha\)-particle of energy 5 MeV is moving forward for a head on collision. The distance of closest approach from the nucleus of atomic number \(\mathrm{Z}=50\) is _________ \(\times 10^{-14} \mathrm{~m}\).
( \(k=9 \times 10^{9} \mathrm{SI}, e=1.6 \times 10^{-19} \mathrm{C} 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\) )
- A 0.72
- B 2.88
- C 1.44
- D 5.76
Answer & Solution
Correct Answer
(B) 2.88
Step-by-step Solution
Detailed explanation
(B) 2.88
Distance of closest approach
\(
\begin{aligned}
r_{0}= & \frac{k q_{1} q_{2}}{\mathrm{~K}} \\
r_{0}= & \frac{9 \times 10^{9} \times(2 e)(\mathrm{Ze})}{5 \times 10^{6} \times 1.6 \times 10^{-19}} \\
& 9 \times 10^{9} \times 2 \times 50 \\
r_{0}= & \frac{\times\left(1.6 \times 10^{-19}\right)^{2}}{5 \times 10^{6} \times 1.6 \times 10^{-19}} \\
r_{0}= & 2.88 \times 10^{-14} \mathrm{~m}
\end{aligned}
\)
Distance of closest approach
\(
\begin{aligned}
r_{0}= & \frac{k q_{1} q_{2}}{\mathrm{~K}} \\
r_{0}= & \frac{9 \times 10^{9} \times(2 e)(\mathrm{Ze})}{5 \times 10^{6} \times 1.6 \times 10^{-19}} \\
& 9 \times 10^{9} \times 2 \times 50 \\
r_{0}= & \frac{\times\left(1.6 \times 10^{-19}\right)^{2}}{5 \times 10^{6} \times 1.6 \times 10^{-19}} \\
r_{0}= & 2.88 \times 10^{-14} \mathrm{~m}
\end{aligned}
\)
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