A) 25%
B) 75%
C) 60%
D) 50%
E) 30%
Correct Answer: B
Solution :
From de-Broglies formula \[\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mK}}\] \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{\sqrt{{{K}_{2}}}}{\sqrt{{{K}_{1}}}}\] Given, \[{{K}_{2}}=16{{K}_{1}}\] \[\therefore \] \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\sqrt{16}=4\] \[\Rightarrow \] \[{{\lambda }_{2}}=\frac{{{\lambda }_{1}}}{4}\] % change\[=\frac{{{\lambda }_{2}}-{{\lambda }_{1}}}{{{\lambda }_{1}}}\times 100\] % change\[=\frac{3}{4}\times 100=75%\]
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