JEE Main 2026 — Jan 21, Shift 1 — Physics solution

JEE Main2026Jan 21, Shift 1Physics

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 $\frac{1}{4 π ϵ _{0}} = 9 \times 1 0^{9}$ in SI units)

A
$2.95 \times 1 0^{- 14}$
B
$2.95 \times 1 0^{- 16}$
C
$3.85 \times 1 0^{- 16}$
D
$3.85 \times 1 0^{- 14}$

Solution

At the distance of closest approach $r$ , all the kinetic energy of the $α$ -particle (charge $+ 2 e$ ) is converted to electrostatic PE with the gold nucleus (charge $+ 79 e$ ):

$K = \frac{1}{4 π ε _{0}} \cdot \frac{( 2 e ) ( 79 e )}{r}$ .

$K = 7.7 \times 1 0^{6} \times 1.6 \times 1 0^{- 19}$ J. Solving:

$r = \frac{( 9 \times 1 0 ^{9} ) ( 2 ) ( 79 ) ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{7.7 \times 1 0 ^{6} \times 1.6 \times 1 0 ^{- 19}} \approx 2.95 \times 1 0^{- 14}$ m.

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