The kinetic energy of an electron in the second Bohr's orbit of a hydrogen atom is [a0 is Bohr's radius]

The kinetic energy of an electron in the nth Bohr orbit is given by:
K.E. = 1/2 mv²
where m is the mass of the electron and v is its velocity.
According to Bohr's model, the velocity of an electron in the nth orbit is given by:
v = v₀/n
where v₀ = e²/2ε₀h
For hydrogen atom (Z=1), v₀ = e²/2ε₀h = 2.18 x 10⁶ m/s
In the second orbit (n=2), v = v₀/2 = (2.18 x 10⁶)/2 = 1.09 x 10⁶ m/s
Therefore, K.E. = 1/2 * m * (v₀/2)² = 1/8 m v₀²
According to Bohr's theory, the radius of the nth orbit is given by:
r = n² a₀
where a₀ is the Bohr radius (a₀ = 0.53 Å)
The kinetic energy can also be expressed in terms of Bohr radius (a₀):
K.E. = 1/2 mv² = (me⁴Z²/8ε₀²h²)(1/n²)
For hydrogen atom (Z=1) and n=2, we have:
K.E. = me⁴/32ε₀²h² = h²/32π²ma₀²
Since a₀ = ε₀h²/πme², we can substitute this into the above expression:
K.E. = h²/32π²m(ε₀h²/πme²)² = h²/(32π²m(ε₀²h⁴/π²m²e⁴)) = me⁴/32ε₀²h²
K.E. = h²/32π²ma₀² ≈ h²/16π²ma₀²
Therefore, the kinetic energy of an electron in the second Bohr's orbit is h²/16π²ma₀² (option a)