A) \[<\,124\,kV\]
B) \[>\,124\,kV\]
C) between 60 kV and 70 kV
D) = 100 kV
Correct Answer: A
Solution :
From conservation of energy the kinetic energy of electron equals the maximum photon energy (we neglect the work function\[\text{o }\!\!|\!\!\text{ }\]because it is normally so small compared to\[\text{e}{{\text{V}}_{0}}\]). \[\therefore \] \[e{{V}_{0}}=h{{v}_{\max }}\] or \[e{{V}_{0}}=\frac{hc}{{{\lambda }_{\min }}}\] \[\therefore \] \[{{V}_{0}}=\frac{hc}{e{{\lambda }_{\min }}}\] or \[{{V}_{0}}=\frac{12400\times {{10}^{-10}}}{{{10}^{-11}}}\] \[=124\,kV\] Hence, accelerating voltage for electrons in X-ray machine should be less than 124 kV.
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