A) A decrease of \[11\text{ }%\]
B) An increase of \[11.1%\]
C) An increase of \[10%\]
D) An increase of \[\text{23}\text{.4}%\]
Correct Answer: D
Solution :
| using the relation, \[\lambda \propto \frac{1}{K.E.},\] |
| If \[{{\lambda }_{1}}=\lambda \]and \[\operatorname{K}.E{{.}_{1}}=E\](initial) |
| Then \[{{\lambda }_{2}}=0.9\lambda \] |
| [\[10%\]Decrease from \[{{\lambda }_{1}},\] |
| \[{{\lambda }_{2}}=\lambda -\frac{10}{100}\lambda \]\[=\lambda \left( 1-0.1 \right)=0.9\lambda ]\] |
| \[\therefore \]\[\operatorname{K}.E{{.}_{2}}=\frac{\lambda _{1}^{2}}{\lambda _{2}^{2}}\times \operatorname{K}.E{{.}_{1}}=\frac{{{\lambda }^{2}}}{{{\left( .9\lambda \right)}^{2}}}\times \operatorname{K}.E{{.}_{1}}\]\[=\frac{{{\lambda }^{2}}\times 100}{81{{\lambda }^{2}}}\times \operatorname{E}\] |
| \[=\frac{100}{81}\operatorname{E}\] [more than\[\operatorname{K}.E{{.}_{1}}i.e.,\] increase] |
| Now increase in \[K.E.=K.E{{.}_{2}}-K.E{{.}_{1}}\]\[a=\frac{100}{81}E-E\] |
| Or \[%\]increase of \[\operatorname{K}.E.=\frac{\frac{100}{81}E-E}{E}\times 100\] \[=\frac{19}{81}\times 100=23.4%\] |
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