If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ______ m (Atomic number of gold = 7

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Solution

Q.
If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ______ m
(Atomic number of gold = 79 and $\frac{1}{4 π ε_{0}} = 9 \times 10^{9}$ in SI units) [2026]

1 $3.85 \times 10^{- 16}$

2 $2.95 \times 10^{- 14}$

3 $3.85 \times 10^{- 14}$

4 $2.95 \times 10^{- 16}$

Ans.
(2)

Energy conservation

$K_{i} + U_{i} = K_{f} + U_{f}$

$7.7 \times 10^{6} \times 1.6 \times 10^{- 19} + 0$

$= 0 + \frac{9 \times 10^{9} (1.6 \times 10^{- 19}) (79 \times 1.6 \times 10^{- 19})}{r}$

$r = 2.95 \times 10^{- 14}$

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