JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Modern Physics

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 ground state to that in excited state is

Answer: (A)

Solution

The de Broglie wavelength of an orbiting electron follows from the Bohr quantisation condition

m v r = \frac{nh}{2 π}

, so

m v = \frac{nh}{2 π r}

and

λ = \frac{h}{m v} = \frac{2 π r h}{nh} \propto \frac{r}{n} .

For the ground state,

n = 1

r_{1} = 5.3 \times 1 0^{- 11} m

; for the third excited state,

n = 4

r_{4} = 8.48 \times 1 0^{- 10} m

. The ratio of the two wavelengths is

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

The wavelengths are therefore in the ratio

1 : 4

, and the value of this ratio is 4.

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Single correct option