Bohr postulates that the angular momentum of an electron in a circular path is quantized and is given by mvr = nh/2π, where n = 0, 1, 2, 3,... (1.c). Find an expression for the radius of their orbit.

To find the expression for the radius of the orbit, we can rearrange the equation mvr = nh/2π to solve for r:

r = nh/2πmv

Since the mass of the electron (m) and Planck's constant (h) are constants, we can simplify the expression further:

r = n(h/2πm)v

We know that the velocity of the electron in a circular orbit is related to the radius of the orbit by the equation v = ωr, where ω is the angular velocity of the electron. The angular velocity can be expressed as ω = v/r.

Substitute v = ωr into the expression for r:

r = n(h/2πm)(ωr)

r = n(h/2πm)ωr

Now, we can solve for the radius of the orbit:

r = n(h/2πm)ωr

r = n(h/2πm)ωr

1 = n(h/2πm)ω

r = h/2πmω

Therefore, the expression for the radius of the orbit is r = h/2πmω.

The angular momentum of an electron in a circular path is given by:

$$mvr=nh/2π$$

where:

- m is the mass of the electron
- v is the speed of the electron
- r is the radius of the orbit
- n is an integer (0, 1, 2, 3, ...)

Solving for the radius of the orbit, we get:

$$r=\frac{nh}{2πmv}$$

This expression gives the radius of the orbit for a given value of n. The radius of the orbit increases as n increases.

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