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Stopping potentials of 24,100,110,115 \mathrm{kV} are measured for photoelectrons emitted from a certain element when it is irradiated with monochromatic \mathrm{X}-rays. The element is used as a target in an \mathrm{X}-ray tube. The energy of \mathrm{\mathrm{K}_\alpha} line is -


 

Option: 1

54 \mathrm{KeV}
 


Option: 2

76 \mathrm{KeV}
 


Option: 3

88 \mathrm{KeV}
 


Option: 4

32 \mathrm{KeV}


Answers (1)

best_answer

Let \mathrm{E_K, E_L, E_M, E_N} be the binding energies of \mathrm{K, L, M \: \: and \: \: N} shell. Let \mathrm{E_P} be energy of incident photon. Then

\mathrm{ E_P-E_K=24 \mathrm{KeV} }  .........(1)

\mathrm{ E_P-E_L=100 \mathrm{KeV} }..........(2)

\mathrm{ E_P-E_M=110 \mathrm{KeV} }........(3)

\mathrm{ E\left(K_\alpha\right)=E_K-E_L=100-24=76 \mathrm{KeV} }

Hence option 2 is correct.




 

Posted by

Pankaj

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