The kinetic energy of an electron in the second
Bohr's orbit of a hydrogen atom is (ao is Bohr's
radius)
h2
4t?m az
(b)
16?ma?
H2
(C) 320?ma
(d)
h2
(d) 64 mma?

Dear Student,

According to Bohr's postulate 

mvr = nh/2$\pi$

squaring both side
m²v²r² = n²h²/ 4$\pi$²
1/2 mv² = n²h²/[8($\pi$²)mr²]
Placing r² = (16a₀)² for n =2

K.E = (2²) h/ [8($\pi$²)m.16a₀]
= h²/[32$\pi$²ma₀]

Answer c) K.E = = h²/[32$\pi$²ma₀]

Regards