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Calculate the accelerating potential that must be imparted to a proton beam to give it an effective wavelength of 0.005nm.

$\begin{array}{1 1}(a)\;20.5V&(b)\;32.85V\\(c)\;33.36V&(d)\;31.4V\end{array}$

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$u=\large\frac{h}{m\lambda}$
Mass of proton=$1.67\times 10^{-27}$
$u=\large\frac{6.625\times 10^{-34}}{1.67\times 10^{-27}\times 0.005\times 10^{-9}}$
$\Rightarrow 7.94\times 10^4m/s$
Accelerating potential is V,then velocity (u) acquired by the charge particle having charge Q and mass m.
$QV=\large\frac{1}{2}$$mu^2$
$u=\sqrt{\large\frac{2QV}{m}}$
$\;\;\;=\sqrt{\large\frac{2\times 1.602\times 10^{-19}\times V}{1.67\times 10^{-27}}}$
$7.94\times 10^4=\sqrt{\large\frac{2\times 1.602\times 10^{-19}\times V}{1.67\times 10^{-27}}}$
$V=32.85V$
Hence (b) is the correct answer.
answered Jan 16, 2014 by sreemathi.v
 

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