A) 12.4 kV
B) 24.8 kV
C) 124 kV
D) 248 kV
Correct Answer: C
Solution :
\[{{\lambda }_{\min }}=\frac{hc}{eV}=\frac{12375}{V}{\AA};\] \[\therefore \,\,\,V=\frac{12375}{\lambda \,\,\text{in}\,\,{\AA}}=124\,\,kV\]
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