A) 105 J
B) 105 MeV
C) 10?1 MeV
D) 105 KeV
Correct Answer: C
Solution :
Since \[{{\lambda }_{\min }}=\frac{12375}{V}{\AA}\]\[=\frac{12375}{{{10}^{5}}}\,{\AA}=0.123\,{\AA}\] \[{{E}_{\max }}=\frac{hc}{{{\lambda }_{\min }}};\] On putting the values. \[{{E}_{\max }}\cong {{10}^{-1}}MeV.\]
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