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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

`lambda=6.6Å`

B

`E=3.4eV`

C

Both are correct

D

Both are wrong

Text Solution

Verified by Experts

The correct Answer is:
C
`KE = |E| = 3.4 eV
`lambda(in Å)` = sqrt((150)/(KE(in eV))`
`sqrt(150)/(3.4)`
=6.6 Å
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