Find the wavelength (in angstrom) of the photon emitted when an electron jumps from the second Bohr orbit to the first bohr orbit of hydrogen atom. Ionisation potential of hydrogen atom in its ground energy state `=2.17xx10^(-11)` erg`*"atom"^(-1)`

Explanation

Energy of the electron in first orbit (n=1) of hydrogen atom, `E_(1)=(2pi^(2)me^(4))/(h^(2))`
Ionisation potential of hydrogen atom. i.e., the energy required to move the electron from n=1 energy level to an infinite distance`=2.17xx10^(-11)`erg.
`therefore`Energy of electron in 1st (n=1) orbit`=-2.17xx10^(-11)` erg
i.e., `-(2pi^(2)me^(4))/(h^(2))=-2.17xx10^(-11)`
Energy of the electron in second orbit (n=2),
`E_(2)=(2pi^(2)me^(4))/(2^(2)h^(2))=(2.17xx10^(-11))/(2^(2))=-0.5425xx10^(-11)`erg
`thereforeDeltaE=E_(2)-E_(1)=[(-0.5425xx10^(-11))-(-2.17xx10^(-11))]`
`=(-0.5425+2.17)xx10^(-11)=1.6275xx10^(-11)` erg
`DeltaE=hv or, DeltaE=hxx(c)/(lamda)`
hence, `1.6275xx10^(-11)=(6.626xx10^(-27)xx3xx10^(10))/(lamda)`
`therefore lamda=(6.626xx10^(-27)xx3xx10^(10))/(1.6275xx10^(-11))`
`=1221xx10^(-8)cm=1221` Å `[because1Å=10^(-8)cm]`.