The de-Broglie wavelength of an electron in the 4 th orbit is \ \ \ \ π a 0 . (a 0 = Bohr radius )

Chemistry · Structure of Atom

JEE Main 2024 — 4 April, Shift 1 — Question 77

The de-Broglie wavelength of an electron in the $4^{\text {th }}$ orbit is $\_\_\_\_ \pi a_0$ .

$(a_0 =$ Bohr radius $)$

Numerical answer — enter this value.

Standing wave condition, we have

2\pi r_n = n\lambda

For $n=4$ , we have

\lambda = \frac{2\pi r_4}{4}

Bohr radius, we have

r_n = n^2 a_0

r_4 = 16a_0

\lambda = \frac{2\pi(16a_0)}{4}

\lambda = 8\pi a_0

Answer key and solution verified before publishing.

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