The difference between the radii of `n^(th) and (n + 1)^(th)` orbits of hydrogen atoms is equal to the radius of `(n-1)^(th)` orbit of hydrogen. The angular momentum of the electron in the `n^(th)` orbit is ___ (h is Planck's constant)

Options

A. `(h)/(pi)`
B. `(2h)/(pi)`
C. `(3h)/(pi)`
D. `(4h)/(pi)`

Answer

B

Explanation

Radius of `n^(th)` orbit in an atom,
`r_(n) = (n^(2)h^(2))/(4pi^(2) mZe^(2))`
So, `r_(n) prop n^(2)`
Given in question, difference between radii of `(n+1)^(th) and n^(th)` orbit = radius of `n^(th)` orbit.
`rArr (n+1)^(2) -n^(2) = (n-1)^(2)`
`n^(2) -4n = 0 or n=4`
According to Bohr.s postulate, angular momentum is an integral multiple of `(h)/(2pi)`.
So, for `4^(th)` orbit,
`L = (nh)/(2pi) = (4h)/(2pi) = (2h)/(pi)`