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A material whose \mathrm{K} absorption edge is \mathrm{0.15 \AA} is irradiated with \mathrm{0.1 \AA} \mathrm{X}-rays. The maximum kinetic energy of photoelectrons that are emitted from \mathrm{K}-shell is -
 

Option: 1

41 \mathrm{KeV}


 


Option: 2

51 \mathrm{KeV}
 


Option: 3

61 \mathrm{KeV}
 


Option: 4

71 \mathrm{KeV}


Answers (1)

best_answer

\left|\mathrm{E}_{\mathrm{K}}\right|=\frac{\mathrm{hc}}{\lambda_{\mathrm{K}}}=\frac{12.4 \mathrm{KeV} \AA}{0.15 \AA}=82.7 \mathrm{KeV}

The energy of incident photon

\mathrm{E_v=\frac{h c}{\lambda}=\frac{12.4}{0.1}=124 \mathrm{KeV}}
The maximum kinetic energy is

\mathrm{ K_{\max }=E_v-\left|E_k\right|=41.3 \mathrm{KeV} \simeq 41 \mathrm{KeV} }

Hence option 1 is correct.

Posted by

Ritika Jonwal

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