An electron in an atom has a speed of 2.2 × 10^6 m/s and orbits the nucleus at a distance of 5 × 10^-11m.a)What is its centripetal acceleration? b)A neutron star of radius 17 km is found to rotate at 5 revolutions per second. What is the centripetal acceleration of a point on its equator?

a) To find the centripetal acceleration of an electron orbiting the nucleus, we can use the formula:

Centripetal acceleration = (velocity)^2 / radius

Given:
Velocity of electron (v) = 2.2 × 10^6 m/s
Radius of orbit (r) = 5 × 10^-11 m

Substituting the given values into the formula:
Centripetal acceleration = (2.2 × 10^6 m/s)^2 / (5 × 10^-11 m)

Using a calculator, we can calculate the centripetal acceleration:

Centripetal acceleration = (2.2 × 10^6)^2 / (5 × 10^-11)
Centripetal acceleration ≈ 193.6 × 10^16 m/s^2 (using scientific notation)

Therefore, the centripetal acceleration of the electron is approximately 193.6 × 10^16 m/s^2.

b) To find the centripetal acceleration of a point on the equator of a rotating neutron star, we can use the formula:

Centripetal acceleration = (Angular velocity)^2 × radius

Given:
Angular velocity (ω) = 5 revolutions per second (we need to convert revolutions to radians)
Radius of star (r) = 17 km = 17 × 10^3 m

To convert revolutions per second to radians per second, we need to multiply by the conversion factor: 2π radians / 1 revolution.

Angular velocity (in radians per second) = 5 revolutions per second × 2π radians / 1 revolution

Substituting the given values into the formula:
Centripetal acceleration = (5 × 2π)^2 × 17 × 10^3

Using a calculator, we can calculate the centripetal acceleration:

Centripetal acceleration = (5 × 2π)^2 × 17 × 10^3
Centripetal acceleration ≈ 53.38 × 10^6 m/s^2 (using scientific notation)

Therefore, the centripetal acceleration of a point on the equator of the neutron star is approximately 53.38 × 10^6 m/s^2.