A) 25
B) 75
C) 60
D) 50
Correct Answer: B
Solution :
[b] \[{{E}_{1}}={{E}_{0}},{{E}_{2}}=16{{E}_{0}}\] \[{{\lambda }_{\,0}}=\frac{h}{4\sqrt{2m}{{E}_{0}}},{{\lambda }_{1}}=\frac{h}{\sqrt{2m\times 16{{E}_{0}}}}\] \[{{\lambda }_{1}}=\frac{h}{4\sqrt{2m{{E}_{0}}}}=\frac{{{\lambda }_{0}}}{4}\] \[%\,change=\frac{{{\lambda }_{0}}-\frac{{{\lambda }_{0}}}{4}}{{{\lambda }_{0}}}\times 100\] \[=75%\]
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