A) \[\frac{0.287}{\sqrt{V}}\overset{0}{\mathop{A}}\,\]
B) \[\frac{12.27}{\sqrt{V}}\overset{0}{\mathop{A}}\,\]
C) \[\frac{0.101}{\sqrt{V}}\overset{0}{\mathop{A}}\,\]
D) \[\frac{0.202}{\sqrt{V}}\overset{0}{\mathop{A}}\,\]
Correct Answer: C
Solution :
de-Broglie wavelength is given by \[\lambda =\frac{h}{\sqrt{2mE}}\] Since, charge on \[\alpha -\]particle =2e and if it is accelerated through V volts, then its K.E. \[E=2\,\,eV\] \[\therefore \] \[\lambda =\frac{h}{\sqrt{2m.2eV}}\] or \[\lambda =\frac{6.63\times {{10}^{-34}}}{\sqrt{4\times 6.6465\times {{10}^{-27}}\times 1.6\times {{10}^{-19}}\times V}}m\] \[=\frac{3.315\times {{10}^{-34}}}{\sqrt{1.6\times 6.6465\times {{10}^{-25}}\sqrt{V}}}m\] \[=\frac{0.3315\times {{10}^{-10}}}{\sqrt{10.6344}}.\frac{1}{\sqrt{V}}m\] \[=\frac{0.101}{\sqrt{V}}\overset{\text{o}}{\mathop{\text{A}}}\,\]
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