Question

The accelerating potential that must be imparted to a proton beam to give it an effective wavelength of $0.05\, nm$ is (wt. of one mole of proton $=1.008 g$ )

• A. $0.325\, eV$
• B. $5.205\, eV$
• C. $52.05\, eV$
• D. $3.25\, eV$

Answer 1

Answer: (A)

Weight of one proton $=\frac{1.008 \times 10^{-5}}{6.023 \times 10^{23}} kg$

We know, $\lambda=\frac{h}{m v}$

$v=\frac{h}{m \lambda}, \lambda=0.05 nm =0.05 \times 10^{-9}$

$ m =5 \times 10^{-11} m$

$K . E .=\frac{1}{2} m v^{2}=\frac{1}{2} m\left(\frac{h}{m \lambda}\right)^{2}$

$=\frac{1}{2} \frac{h^{2}}{m \lambda^{2}}$

$=\frac{1}{2} \times \frac{\left(6.6 \times 10^{-34}\right)^{2} \times 6.023 \times 10^{23}}{1.008 \times 10^{-3} \times\left(5 \times 10^{-11}\right)^{2}}$

$=5.205 \times 10^{-20} J$

$=\frac{5.205 \times 10^{-20}}{1.6 \times 10^{-19}} eV =3.25 \times 10^{-1}$

$=0.325\, eV$