13.6eV is needed for ionisation of a hydrogen atom. An electron in a hydrogen atom in its ground state absorbs 1.50 times as much energy as the minimum energy required for it to escape ffrom the atom. What is the wavelength of the emitted electron? (m_(e)= 9.109 xx 10^(-31)kg, e= 1.602 xx 10^(-19) coulomb, h= 6.63 xx 10^(-34) J.s)

13.6eV is needed for ionisation of a hydrogen atom. An electron in a h

1.	13.6eV is needed for ionisation of a hydrogen atom. An electron in a hydrogen atom in its ground state absorbs 1.50 times as much energy as the minimum energy required for it to escape ffrom the atom. What is the wavelength of the emitted electron? (m_(e)= 9.109 xx 10^(-31)kg, e= 1.602 xx 10^(-19) coulomb, h= 6.63 xx 10^(-34) J.s)
Answer» SOLUTION :1.5 times of 13.6eV, i.e., 20.4eV, is absorbed by the hydrogen atom out of which 6.8eV (20.4-13.6) is CONVERTED to kinetic energy. KE = 6.8eV= 6.8 (`1.602 xx 10^(-19)` coulomb)(1 volt)= `1.09 xx 10^(-18)J`. Now, KE `=(1)/(2) MV^(2)` or `v= sqrt((2KE)/(m))= sqrt((2(1.09 xx 10^(-18)J))/((9.109 xx 10^(-31)kg)))= 1.55 xx 10^(6) m//s` `therefore lamda= (h)/(mv)= ((6.63 xx 10^(-34) J.s))/((9.109 xx 10^(-31)kg) (1.55 xx 10^(6) m//s))` `=4.70 xx 10^(-10)` metres.