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 x `10^(-11)` erg per atom.

Answer

`1220A^(@)`

Explanation

`overset(DeltaE)(2rarr1)=I.E.[(1)/(4)-(1)/(1)]`
`2.17xx10^(-11)erg//"atom"[(1)/(4)-(1)/(1)]=(hc)/(lambda(m))`
`2.17xx10^(-11)xx10^(-7)[(1)/(4)-(1)/(1)]=(6.626xx10^(-34)xx3xx10^(8))/(lambda)`
`lambda=(6.626xx10^(34)xx3xx10^(8)xx4)/(2.17xx10^(-18)xx3)=(6.626xx4xx10^(8))/(2.17)`
`=12.20xx10^(-8)m`
`1mrarr10^(10)Å`
`6.10xx10^(-8)mrarr(12.2xx10^(10))/(10^(8))=1220Å`