Question

What is the de-Broglie wavelength of the $ \alpha $ -particle accelerated through a potential difference of V volt? (mass of $ \alpha $ particle $ =6.6465\,\times {{10}^{-27}}\,kg $ )

• 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}}\, $

Answer 1

Answer: (C)

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}}}\, $