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When an electron in an excited state of ...

When an electron in an excited state of Mo atom falls L to K -shell, an X -ray is emitted. These X -rays are diffranted at angle of ` 7.75^(@)` by planes with a sepration of 2.64Å . What is the difference in energy between K-shelll and L -shell in Mo, assuming a first order diffraction ? ` ( sin 7.75^(@) = 0.1349)

A

`36.88 xx 10^(-19) J`

B

`27.88 xx 10^(-16)` J

C

`63.88 xx 10^(-17) J`

D

`64.88 xx 10^(-16)` J

Text Solution

Verified by Experts

The correct Answer is:
B
`2d sin theta = n lambda`
`lambda = 2d sin theta = 2 xx 2.64 xx 10^(-10) xx sin 7.75^(@)`
`=0.7123 xx 10^(-10) m`
`E = (hc)/lambda= (6.62 xx 10^(-34) xx 3 xx 10^(8))/(0.7123 xx 10^(-10)) = 27.88 xx 10^(-16) J`
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