A) must be increased by \[\frac{100}{3}%\]
B) must be decreased by \[\frac{100}{3}%\]
C) must increased by 25%
D) must decreased by 25%.
Correct Answer: A
Solution :
[a] \[{{\lambda }_{1}}=\frac{hc}{e{{v}_{1}}};{{\lambda }_{2}}=\frac{hc}{e{{v}_{2}}}\Rightarrow \frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{{{v}_{1}}}{{{v}_{2}}}\] \[\Rightarrow \frac{3}{4}=\frac{{{v}_{1}}}{{{v}_{2}}}\Rightarrow \left( \frac{{{v}_{2}}-{{v}_{1}}}{{{v}_{1}}} \right)\times 100=\frac{100}{3}\] Hence, P.D. must be increased by \[\frac{100}{3}%.\]You need to login to perform this action.
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