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 \[\phi \] because it is normally so small compared to \[e{{V}_{0}})\]. \[\therefore \] \[e{{V}_{0}}=h{{v}_{\max }}\] or \[e{{V}_{0}}=\frac{hc}{{{\lambda }_{\min }}}\] \[\therefore \] \[{{V}_{0}}=\frac{hc}{e{{\lambda }_{\min }}}\] or \[{{V}_{0}}=\frac{12400\times {{10}^{-10}}}{{{10}^{-11}}}\] Hence, accelerating voltage for electrons in \[X-\]ray machine should be less than\[124\,\,kV\].
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