A) one fourth
B) half
C) four times
D) two times
Correct Answer: B
Solution :
de - Broglie wavelength is given by : \[\lambda =\frac{h}{mv}\] ...(i) \[K.E.=\frac{h}{2}m{{v}^{2}}\] \[{{v}^{2}}=\frac{2KE}{m}\] \[v=\sqrt{\frac{2KE}{m}}\] Substituting this in equation (i) \[\lambda =\frac{h}{m}\sqrt{\frac{m}{2KE}}\] \[\lambda =h\sqrt{\frac{1}{2m(K.E.)}}\]i.e.\[\lambda =\propto \frac{1}{\sqrt{KE}}\] \[\therefore \]when KE become 4 times wavelength become 1/2.
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