1.

If the total energy of an electron in a hydrogenatom in excited state is - 3.4 eV, 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 RARR -3.4 eV = -13.6/n^2 rArr n^2= 13.6/3.4 =4`
`therefore` n=2
VELOCITY of electron in n =2 is
`V=(2.19xx10^6)/2 m s^(-1)`
`therefore lambda=h/(mv)=(6.6xx10^(-34)xx2)/(9.1xx10^(-31)xx2.19xx10^6)=6.6xx10^(-10)` m


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