Atomic Structure JEE Main Solution | de Broglie Wavelength

JEE MAIN 2025

29-01-2025 SHIFT-1

Question

If ${a_0}$ is denoted as the Bohr radius of hydrogen atom, then what is the de-Broglie wavelength $(\lambda )$ of the electron present in the second orbit of hydrogen atom? [n : any integer]

Select the correct option:

$\frac{{2{{\rm{a}}_0}}}{{{\rm{n}}\pi }}$

$\frac{{8\pi {a_0}}}{n}$

$\frac{{4{\rm{n}}}}{{\pi {{\rm{a}}_0}}}$

$\frac{{4\pi {a_0}}}{n}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$2\pi {{\rm{r}}_{\rm{n}}} = {\rm{n}}\lambda $
$2\pi \left( {4{{\rm{a}}_0}} \right) = {\rm{n}}\lambda $
$ = \lambda = \frac{{8\pi {{\rm{a}}_0}}}{{\rm{n}}}$

Question Tags

JEE Main

Chemistry

Easy

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