A) 25
B) 75
C) 60
D) 50
Correct Answer: A
Solution :
| For de-Broglie wavelength |
| \[{{\lambda }_{1}}\frac{h}{p}=\frac{h}{\sqrt{2mK}}\,\] .(i) |
| \[{{\lambda }_{2}}=\frac{h}{\sqrt{2m16K}}=\frac{h}{4\sqrt{2mK}}=\frac{{{\lambda }_{1}}}{4}\] (ii) |
| \[{{\lambda }_{2}}=25%\] of \[{{\lambda }_{1}}\] |
| There is 75% change in the wavelength. |
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