Question

Calculate the wavelength of photon which will be emitted when the electron of hydrogen atom jumps from fourth shell to the first shell. Ionisation energy of hydrogen atom is `1.312xx10^(3)hJ "mol"^(-1)`.

Answer 1

Correct Answer - `9.725xx10^(-8)m`
Step I. Amount of energy released
lonisation energy of hydrogen atom `=1.312xx10^(3) kJ "mol"^(-1)`
`=(1.312xx10^(3)xx10^(3))/(6.022xx10^(23))=2.18xx10^(-18) J "atom"^(-1)`
Energy of electron in first shell `(E_(1))=E_(prop)-l.E.`
`=0-2.18xx10^(-18) J "atm"^(-1)=-2.18xx10^(-18) J "atom"^(-1)`
Energy of electron in fourth shell `(E_(4))=(-2.18xx10^(-18))/(n^(2))=(-2.18xx10^(-18))/((4)^(2))`
`=0.136xx10^(-18) J "atom"^(-1)`
Amount of energy released `(Delta E)=E_(4)-E_(1)=(-0.136xx10^(-18))-(-2.18xx10^(-18))`
`=2.044xx10^(-18) J "atom"^(-1)`
Step II. Wavelength of photon emitted
Energy (E)=`hv=(hc)/(lambda) , lambda=(hc)/(E)=((6.626xx10^(-34)Js)(3xx10^(8)ms^(-1)))/((2.044xx10^(-18) J))=9.725xx10^(-8)m`.