Hydrogen is the simplest atom of nature. There is one proton in its nucleus and an electron moves around the nucleus in a circular orbit. According to Niels Bohr's, this electron moves in a stationary orbit, if emits no electromagnetic radiation. The angular momentum of the electron is quantized , i.e., `mvr = (nh//2 pi)`, where `m =` mass of the electron , `v =`velocity of the electron in the orbit , `r =`radius of the orbit , and `n = 1, 2, 3`.... When transition takes place from `Kth` orbit to `Jth` orbit, energy photon is emitted. If the wavelength of the emitted photon is `lambda`.

we find that `(1)/(lambda) = R [(1)/(J^(2)) - (1)/(K^(2))]`, where `R` is
Rydberg's constant.
On a different planed, the hydrogen atom's structure was somewhat different from ours. The angular momentum of electron was `P = 2n (h//2 pi)`. i.e., an even multipal of `(h//2 pi)`.
Answer the following questions regarding the other planet based on above passage:
The minimum permissible radius of the orbit will be

Answer

B

Explanation

On other planet: `mvr = 2n (h)/(2 pi) implies v = (nh)/(pi mr)`
`(mv^(2))/( r) = (1)/(4 pi epsilon_(0)) (e^(2))/( r^(2)) implies (m n^(2) h^(2))/(n^(2) m^(2) r^(3)) = (1)/(4 pi epsilon_(0)) (e^(2))/( r^(2))`
Putting `n = 1, we get r = (4 pi epsilon_(0))/(m pi e^(2))`