Question

angular momentum of an electron revolving in bohr's orbit is 2h/pi radius of this bohr's orbit is: a),0.529 angstrom b)1.058 angstrom c)2.116 angstorm d)8.464 angstorm

Answer 1

Explanation:

.vr=nh/2pi=2h/pi

n/2=2

n=4

r (nth orbit)=0.532n^2/z A°

=0.532×16 A°

=8.464A°●[APPROXIMATELY]●~~~~~~~~~~

Answer 2

Option D is the right answer. angular momentum of an electron revolving in Bohr's orbit is 2h/pi, and the radius of this Bohr's orbit is 8.464 angstrom.

Given:

angular momentum of an electron revolving in Bohr's orbit = 2h/pi

Find:

The radius of the Bohr's orbit.

Solution:

we know, the equation for the angular momentum of Bohr's orbit is = $\frac{nh}{2\pi }$

$\frac{nh}{2\pi }$ = $\frac{2h}{\pi }$

So n = 4

Equation for Bohr's orbit

R = 0.537n²

  = 0.537*16

  ≈ 8.464 angtrom

∴ The radius of the Bohr orbit is 8.464 A⁰