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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:
A
$\frac{{2{{\rm{a}}_0}}}{{{\rm{n}}\pi }}$
B
$\frac{{8\pi {a_0}}}{n}$
C
$\frac{{4{\rm{n}}}}{{\pi {{\rm{a}}_0}}}$
D
$\frac{{4\pi {a_0}}}{n}$
✓ 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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