JEE Main 2026 — Jan 22, Shift 1 — Physics solution

JEE Main2026Jan 22, Shift 1Physics

7.9 MeV $α$ –particle scatters from a target material of atomic number 79. From the given data the estimated diameter of nuclei of the target material is (approximately) ________ m.

$[\frac{1}{4 π ϵ _{0}} = 9 \times 1 0^{9} Nm^{2} / C^{2} and electron charge = 1.6 \times 1 0^{- 19} C]$

A
$5.76 \times 1 0^{- 14}$
B
$1.44 \times 1 0^{- 13}$
C
$2.88 \times 1 0^{- 14}$
D
$1.69 \times 1 0^{- 12}$

Solution

At the distance of closest approach $r$ , all the alpha’s kinetic energy converts into electrostatic potential energy with the target nucleus (charge $Z e$ , $Z = 79$ ). The alpha carries charge $2 e$ .

Energy balance: $K E_{α} = \frac{k ( 2 e ) ( Z e )}{r}$ , so $r = \frac{k ( 2 e ) ( Z e )}{K E _{α}}$ .

Plug numbers (with $K E_{α} = 7.9 \times 1 0^{6} eV = 7.9 \times 1 0^{6} \times 1.6 \times 1 0^{- 19}$ J):

$r = \frac{9 \times 1 0 ^{9} \times 2 \times 79 \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{7.9 \times 1 0 ^{6} \times 1.6 \times 1 0 ^{- 19}} = 2.88 \times 1 0^{- 14}$ m.

Diameter $D = 2 r = 5.76 \times 1 0^{- 14}$ m.

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