Calculate the wavelength in Angstroms of the photon that is emitted when an electron in the Bohr's orbit `n = 2`, returns to the orbit `n = 1`, in the hydrogen atom .The ionisation potential of the ground state hydrogen atom is `2.17 xx 10^(-11)` ergs per atom

Explanation

Since the hydrogen atom has only one orbit containing only one electron, the ionisation potential of the ground state of the hydrogen atom is the energy of the electron of the first orbit, i.e,
`E_(1)= -2.17 xx 10^(-11)` erg
Thus, `E_(2)= (E )/(n^(2))`
`= -(2.17 xx 10^(-11))/(2^(2))`
`therefore` energy of the radiation emitted, `DeltaE= E_(2)- E_(1)`
`= - (2.17 xx 10^(-11))/(2^(2)) -(-2.17 xx 10^(-11))`
`=1.627 xx 10^(-11)` erg
We know `DeltaE= hv= (hc)/(lamda)`
Thus `(hc)/(lamda)= 1.627 xx 10^(-11)`
`lamda= (6.62 xx 10^(-27) xx 3 xx 10^(10))/(1.627 xx 10^(-11)) = 1.22 xx 10^(-5)cm`
`=1220 Å (1 Å= 10^(-8)cm)`