IIT JEE Advanced 2012 Paper 1

According to Bohr’s postulate, the angular momentum of an electron is quantised, that is,
mvr =

nh

2π

⇒ v =

nh

2πmr

Now, substituting the above expression in the kinetic energy expression (

1

2

)mv2, we get
K.E. =

1

2

m(

nh

2πmr

)2 (1)
where r is the radius of Bohr’s orbit, that is, r = a0×n2∕Z. For the second Bohr’s orbit of a hydrogen atom, substituting Z = 1 and n = 2, we get r = 4a0 where a0 is a constant. Now, substituting this value of r in Eq. (1), we get
K.E. =

1

2

×m(

4h2

4π2m2.16a02

) =

h2

32π2ma02