A) \[\text{124}\,\text{kV}\]
B) \[\text{124}\,\text{kV}\]
C) between 60 \[\text{kV}\]and 70\[\text{kV}\]
D) \[=100\,\text{kV}\]
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 \[{{V}_{0}}=\frac{12400\times {{10}^{-10}}}{{{10}^{-11}}}\] \[=124\,kV\] Hence, accelerating voltage for electrons in X-ray machine should be less than 124 kV.
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