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