Determine the number of revolutions made by an electron in one second in the 2^nd Bohr orbit of
H-atom.

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$8.2 \times 10^{14}$

$6.2 \times 10^{14}$

$4.1 \times 10^{12}$

$1.6 \times 10^{10}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

n= number of revolution per second= $\frac{v e l o c i t y o f e l e c t r o n}{c i r c u m f e r e n c e o f o r b i t}$
$V_{n} = 2.188 \times 10^{6} \times \frac{z}{n}, μ_{n} = 0.529 A \times \frac{n^{2}}{z}$
Putting z = 1, and n = 2,
$V_{2} = 2.188 \times 10^{6} \times \frac{1}{2}, μ_{2} = 0.529 \times 4 \times 10^{- 10} = 2.11 \times 10^{- 10}$
$n \frac{1.094 \times 10^{6}}{2 \times 3.14 \times 2.116 \times 10^{- 10}} = 0.082 \times 10^{16}$
$= 8.2 \times 10^{14}$ revolution /sec