Find out the wavelength of the electron orbiting in the ground state of hydrogen atom .

Explanation

As per quantum concept, for a stable nthe orbit of hydrogen atom the total length of orbital path (i.e., the circumference of the orbit) is equal to n times the de - Broglie wavelenght of electron. Thus ,
`2 pi rn = n lambda_("electron")`
For an electron orbiting in the ground state of hydrogen atom , n =1 . Hence , we have
` 2 pi r_(1) = lambda_("electron")`