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Q. The de-Broglie’s wavelength of electron present in first Bohr orbit of ‘H’ atom is :

JEE MainJEE Main 2018Structure of Atom

Solution:

Bohr's radius of first stationary state is $r=0.529 \, \mathring{A}$

Angular momentum is given by $mvr=\frac{n h}{2 \pi}$

For first Bohr's orbit $n=1$

$2 \pi r =\frac{h}{m v}=\lambda $

$\lambda =2 \pi \times 0.529 \, \mathring{A}$