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The kinetic energy of the electron in th...

The kinetic energy of the electron in the second Bohr's orbit of a hydrogen atom [`a_(0)` is Bohr's radius] is

A

`h^(2)/(4pi^(2)ma_(0)^(2))`

B

`(h^(2))/(16pi^(2)ma_(0)^(2))`

C

`h^(2)/(32pi^(2)ma_(0)^(2))`

D

`h^(2)/(64pi^(2) ma_(0)^(2))`

Text Solution

Verified by Experts

The correct Answer is:
C
As per Bohr’s postulate,
Kinetic energy of an electron in 2nd orbit `=((Ze)e)/(2 xx r_(2)), [tau_(n)=(a_(0) xx n^(2))/(Z)], tau_(n)=" radius of n"^(th) shell`
For Hydrogen atom, Z= 1
Kinetic energy in 2nd orbit `=(e^(2))/(2r_(2)) (therefore r_(2)=a_(0) xx (2)^(2)) =(e^(2))/(2 xx a_(0) xx (2)^(2)) =e^(2)/(8a_(0))`
Since `a_(0)=h^(2)/(4pi^(2) me^(2)) therefore " Kinetic energy in II orbit "=h^(2)/(4pi^(2) me^(2) a_(0)) xx (1)/(8a_(0)) =h^2/(32 pi ma_(0)^(2))`
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