A) \[\approx \,\,10\text{ }kV\]
B) \[\approx \,\,16\text{ }kV\]
C) \[\approx \,\,50\text{ }kV\]
D) \[\approx \,\,75\text{ }kV\]
Correct Answer: B
Solution :
\[{{\lambda }_{min}}\,=\,\,\frac{hc}{eV}\,\,\,\,\Rightarrow \,\,{{\lambda }_{1}}\,=\,\,\frac{hc}{e{{V}_{1}}}\,\,and\,\,{{\lambda }_{2}}\,=\,\,\frac{hc}{e{{V}_{2}}}\] \[\therefore \,\,\Delta \lambda ={{\lambda }_{2}}-{{\lambda }_{1}}\,=\,\,\frac{hc}{e}\left[ \frac{1}{{{V}_{2}}}-\frac{1}{{{V}_{1}}} \right]\] \[Given\,\,{{V}_{2}}=1.5\,{{V}_{1}}\] On solving, we get \[{{\operatorname{V}}_{1}} =16000 volt = 16 kV.\]
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