1.

If the total energy of anelectron in a hydrogen like atom in excited state is -3.4eV,then the de Broglie wavelength of the electron is

Answer»

`6.6xx10^(-12)m`
`3xx10^(-10)m`
`6.6xx10^(-10)m`
`9.3xx10^(-12)m`

Solution :`E_(n)=(E_(1))/(n^(2))`
`-3.4eV=-13.6/(n^(2))` or `n^(2)=13.6/3.4=4`,
`:.n=2`
VELOCITY of ELECRON isn=2 is `(v_(1))/2`
`=(2.19xx10^(6))/2ms^(-1)`
`:.lamda=h/(mv)=(6.6xx10^(-34)xx2)/(9.1xx10^(-31)xx2.19xx10^(6))`
`=6.6xx10^(-10)m`


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