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The magnetude of the P.E. of the electro...

The magnetude of the P.E. of the electron in the first orbit of the Bohr's atom is E. Its K.E. is

A

E

B

2E

C

E/2

D

E/4

Text Solution

Verified by Experts

The correct Answer is:
C
P.E. =2 KE
K.E. `=(PE)/2=E/2`
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