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In an excited state of hydrogen like ato...

In an excited state of hydrogen like atom an electron has total energy of `-3.4 eV`. If the kinetic energy of the electron is E and its de-Broglie wavelength is `lambda`, then

A

`E =6.8e V lambda=6.6 xx 10^(-10)m`

B

`E= 3.4eV,lambda=6.6 xx 10^(-10)m`

C

`E=3.4eV,lambda=6.6xx10^(-11)m`

D

`E=6.8eV,lambda=6.6xx10^(-11)m`

Text Solution

Verified by Experts

The correct Answer is:
B
`|T.E|=K.E,lambda=(h)/(sqrt(2mK.E.))`
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