Consider a hydrogen atom in which an electron is revolving in n-1 orbit with 6.57×1015 revolutions per second. Calculate the magnetic moment of the electron.Answer Choices:a 9.2×10 A-m2b. 6.5x10 A-m²c. 5.5×10 A m²d. 7.4x10 A-m²

To solve this problem, we need to use the formula for the magnetic moment of an electron in an atom, which is given by:

μ = e * v * r / 2

where:

• e is the charge of the electron (1.6 * 10^-19 C),
• v is the velocity of the electron, and
• r is the radius of the orbit.

We know that the velocity of the electron can be calculated using the formula:

v = 2 * π * r * n

where n is the number of revolutions per second. We can substitute this into the formula for the magnetic moment to get:

μ = e * 2 * π * r * n * r / 2 = e * π * r^2 * n

We are given that n = 6.57 * 10^15 revolutions per second. We also know that for a hydrogen atom, the radius of the n-1 orbit is given by:

r = 0.529 * n^2 Å

where n is the principal quantum number. For the n-1 orbit, n = 2, so:

r = 0.529 * 2^2 Å = 2.116 Å = 2.116 * 10^-10 m

Substituting these values into the formula for the magnetic moment gives:

μ = 1.6 * 10^-19 C * π * (2.116 * 10^-10 m)^2 * 6.57 * 10^15 s^-1 = 1.6 * π * 4.47 * 10^-20 m^2 * 6.57 * 10^15 s^-1 = 1.6 * π * 4.47 * 6.57 * 10^-5 A m^2 = 37.4 * 10^-5 A m^2 = 3.74 * 10^-4 A m^2

So, none of the answer choices match the calculated value. There might be a mistake in the problem or the answer choices.