Q.

Number of revolutions made by an electron in Bohr’s 2nd orbit of hydrogen atom in one second is (approx)

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

a

8.3×1014

b

6.66×1015

c

5.3×106

d

4.5×1012

answer is A.

(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

[Formula]: Frequency =V2πr=2.18×106×Zn2π×0.529×n2Z=2.18×1062×n3×0.529=2.18×106×12×3.14×23×0.529=8.3×1014(or)[Formula]: Orbital frequency =Z2n3×6.66×1015=18×6.66×1015=8.3×1014
Watch 3-min video & get full concept clarity
Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
whats app icon