Determine the value of the de Broglie wavelength associated with the electron orbiting in the ground state of hydrogen atom ( given `E_(n) = - (13.6//n^(2))eV` and bohr radius `r^(0) = 0.53Å`). How will the de Broglie wavelength change when it is in the first excited state?

Explanation

We know, de Broglie wavelength, ltbgt `lamda = (h)/(mv) = (hr)/(mvr) = (hr)/((nh)/(2pi)) = (2pir)/(n)`
For ground state, n =1
`:. lamda_(1) = (2pi r_(0))/(1) = 2pi xx 0.53 xx 10^(-10)`
`= 3.3 xx 10^(-10) m = 3.3 Å`
Again, `lamda_(n) = (2pi r_(n))/(n) = (2pi n^(2) r_(0))/(n) = n2 pi r_(0) = nlamda_(1)`
When n = 2 (for first excited state)
`lamda_(2) = 2lamda_(1)`
Hence, de Broglie wavelength doubles in the first excited state.