Calculate the de-Broglie wavelength of the electron orbiting in the n= 2 state of hydrogen atom.

Explanation

In n - 2 state of hydrogen atom, as per Bohr.s quantum condition, we have
`mvr_(2) = (nh)/(2pi) = (4h)/(2pi) = (h)/(pi) rArr mv = (h)/(pi r_(2))`
`therefore ` de- Broglie wavelenght ` lambda = (h)/(mv) = (h)/((h//pi r_(2))) = pi r_(2)` , where `r_(2)` is the radius of orbit correspoding to n = 2 state .