CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [ao is radius of first Bohr orbit]:

A
h24π2mao2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
h216π2mao2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
h232π2mao2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
h264π2mao2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C h232π2mao2
As per Bohr's postulate,
mvr=nh2π
So,
v=nh2πmr
KE=12mv2
So,
KE=12m[nh2πmr]2
Since r=ao×n2Z
For 2nd Bohr orbit,
r=ao×221=4ao
KE=12m[22h24π2m2×(4ao)2]=h232π2mao2

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon