1.

Energy of the electron in hydrogen atom is 1.5 timesas much ass the minimum energy required for its escape (13.6eV) from the atom. Wavelength of the emitted electron is

Answer»

`3.96Å`
`5.32Å`
`4.60Å`
`4.71Å`

SOLUTION :Electron in the ground state ABSORBS inerg and the difference in energy provides K.E. to the EMITTED electron.
`E_(1)=13.6eV`
`E_(2)=13.6xx1.5=20.4eV`
`DeltaeE=K.E. =20.4-13.6=6.8xx1.6xx10^(-19)J`
`lamda=h/(sqrt(2mKE))`
`=(6.61xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx6.8xx1.6xx10^(-19)))`
`=4.71xx10^(-10)m=4.71Å`


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