JEE Main 2025 Jan 22 Shift 1, Physics Q7: Wave Nature of Matter

JEE Main2025Jan 22, Shift 1Physics

An electron in the ground state of the hydrogen atom has the orbital radius of $5.3 \times 1 0^{- 11}$ m while that for the electron in third excited state is $8.48 \times 1 0^{- 10}$ m. The ratio of the de Broglie wavelengths of electron in the excited state to that in ground state is

A
4
B
9
C
3
D
16

Solution

From the Bohr quantisation $m v r = \frac{nh}{2 π}$ and $λ = \frac{h}{m v}$ , we get:

$λ = \frac{2 π r}{n}$ , i.e. $λ \propto \frac{r}{n}$ .

Third excited state corresponds to $n = 4$ . So

$\frac{λ _{1}}{λ _{4}} = \frac{r _{1} /1}{r _{4} /4} = \frac{r _{1} \times 4}{r _{4}} = \frac{5.3 \times 1 0 ^{- 11} \times 4}{8.48 \times 1 0 ^{- 10}} = \frac{21.2}{84.8} = \frac{1}{4}$ .

$∴ \frac{λ _{4}}{λ _{1}} = 4$

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JEE Main 2025 Jan 22 Shift 1, Physics Q7: Wave Nature of Matter

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