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. $\frac{0.287}{\sqrt{V}}\stackrel{0}{A}$
• B. $\frac{12.27}{\sqrt{V}}\stackrel{0}{A}$
• C. $\frac{0.101}{\sqrt{V}}\stackrel{0}{A}$
• D. $\frac{0.202}{\sqrt{V}}\stackrel{0}{A}$

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. $E=2eV$ $\therefore$ $\lambda =\frac{h}{\sqrt{2m.2eV}}$ or $\lambda =\frac{6.63\times {10}^{-34}}{\sqrt{4\times 6.6465\times {10}^{-27}\times 1.6\times {10}^{-19}\times V}}m$ $=\frac{3.315\times {10}^{-34}}{\sqrt{1.6\times 6.6465\times {10}^{-25}\sqrt{V}}}m$ $=\frac{0.3315\times {10}^{-10}}{\sqrt{10.6344}}.\frac{1}{\sqrt{V}}m$ $=\frac{0.101}{\sqrt{V}}\stackrel{\text{o}}{A}$