Velocity of an electron in the ground state of a hydrogen atom:
The velocity of an electron in the ground state of a hydrogen atom can be calculated using the Bohr model. According to the Bohr model, the electron orbits the nucleus in a circular path, and the centripetal force required to keep the electron in its orbit is provided by the electrostatic attraction between the electron and nucleus.

The centripetal force can be equated to the electrostatic force using the formula:

mv^2/r = ke^2/r^2

Where:
m is the mass of the electron
v is the velocity of the electron
r is the radius of the orbit
k is the electrostatic constant
e is the charge of the electron

In the ground state of a hydrogen atom, the radius of the orbit can be calculated using the Bohr radius formula:

r = (4πε₀ħ²)/(me²)

Where:
ε₀ is the permittivity of free space
ħ is the reduced Planck's constant
me is the mass of the electron
e is the charge of the electron

By substituting the value of r into the equation for the centripetal force, we can solve for the velocity:

mv^2/r = ke^2/r^2

v² = ke²/mr

v = √(ke²/mr)

Substituting the values of k, e, m, and r, we get:

v = √((9×10^9 Nm²/C²)(1.6×10^-19 C)²/(9.11×10^-31 kg)(5.29×10^-11 m))

v ≈ 2.18×10^6 m/s

Velocity of an electron in the ground state of an If-like atom:
To find the velocity of an electron in the ground state of an If-like atom with a reduced mass of 200 m, we can use the same equation as above, but substitute the reduced mass for the electron mass:

v = √((9×10^9 Nm²/C²)(1.6×10^-19 C)²/(200)(9.11×10^-31 kg)(5.29×10^-11 m))

v ≈ 2.18×10^8 m/s

Therefore, the velocity of an electron in the ground state of an If-like atom with a reduced mass of 200 m is approximately 2.18×10^8 m/s.

Answer: b) 2.18×10^8 m/s (20000vH)