A)
B)
C) Both a and b
D) None of the above
Correct Answer: C
Solution :
As, \[\lambda =\frac{h}{mv}=\frac{h}{p}\] ... (i) and \[K=\frac{1}{2}m{{v}^{2}}=\frac{{{p}^{2}}}{2m}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p=\sqrt{2mK}\] ? (ii) So, from Eqs. (i) and (ii), we get \[\lambda =\frac{h}{\sqrt{2mK}}\,\,\,\Rightarrow \,\,\,\lambda \propto \frac{1}{\sqrt{K}}\] So, graph between \[\lambda \,\,vs\,\,\sqrt{K}\] will be rectangular hyperbola. and \[\lambda \,\,vs\,\,\frac{1}{\sqrt{K}}\] will be straight line.
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