Question

The wavelength of de broglie waves associated with a beam of protons of kinetic energy $5\times {10}^2$eV.
(Mass of each photon$=1.67\times {10}^{-27}$Kg, $h=6.62\times {10}^{-34}$Js.)

• A. $2.42\times {10}^{-12}$m
• B. $4.24\times {10}^{-12}$m
• C. $1.82\times {10}^{-12}$m
• D. $1.28\times {10}^{-12}$m

Answer 1

Answer: (D)

$\frac{1}{2}m{v}^2=E$

$mv=\sqrt{2Em}$

$=\sqrt{2\times 5\times {10}^2\times 1.6\times {10}^{-19}\times 1.6\times {10}^{-27}}$

$=51.69\times {10}^{23}N$

$\lambda =\frac{h}{mv}$

$=\frac{6.62\times {10}^{-34}}{51.63\times {10}^{-23}}$

$=1.28\times {10}^{-12}m$

So, the answer is option (D).