NEET 2026 — Physics solution

NEET2026Physics

In the first excited state of hydrogen atom, the energy of its electron is $- 3.4$ eV. The radial distance of the electron from the hydrogen nucleus in this case is approximately :

(Take 1 eV = $1.6 \times 1 0^{- 19}$ J, e = $1.6 \times 1 0^{- 19}$ C and $\frac{1}{4 π ϵ _{0}} = 9 \times 1 0^{9}$ N m $^{2}$ /C $^{2}$ )

A
$2.1 \times 1 0^{- 8}$ m
B
$2.1 \times 1 0^{- 10}$ m
C
$2.1 \times 1 0^{- 11}$ m
D
$2.1 \times 1 0^{- 9}$ m

Solution

For a hydrogen atom, the total energy in a Bohr orbit relates to the radius by

$∣ E ∣ = \frac{k e ^{2}}{2 r}$

So $r = \frac{k e ^{2}}{2∣ E ∣} = \frac{9 \times 1 0 ^{9} \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{2 \times 3.4 \times 1.6 \times 1 0 ^{- 19}}$

$r \approx 2.1 \times 1 0^{- 10}$ m.

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