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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 = 3.4 eV`

C

both are correct

D

both are wrong

Text Solution

Verified by Experts

The correct Answer is:
C
T.E. `= -3.4 eV`
K.E. `= |T.E.|=3.4 eV`
`lambda_(e)= (12.27)/(sqrt(V))Å=(12.27)/(sqrt(3.4))`
`lambda_(e) ~~ 6.6Å`
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