Question

A Proton and an $/alpha$ particle are accelerated through a potential difference of 100V. The ratio of the wavelength associated with the proton to that associated with an $/alpha$ particle is

• A. $1:2$
• B. $2:1$
• C. $22:1$
• D. $2\sqrt{2}:1$

Answer 1

The correct option is D $2\sqrt{2}:1$

$Enrgy=\frac{Momentu{m}^2}{2m}=p^2/2m$

So momentum $p=\sqrt[2]{22mE}$, $m$ is mass.

Wavelength$\lambda =\frac{h}{p}i.e$ $\lambda \propto \frac{1}{p}$

So required ratio is $\sqrt[2]{2\frac{m_a{E}_a}{m_p{E}_p}}$

Mass of alpha particle is $m_a=4m$ where $m=m_p$ is mass of proton.

Energy of alpha particle $E_a=q_{a}V=2e\times 100$

and energy of proton is $E_p=q_{p}V=e\times 100$

Putting all the values we get the ratio as $\sqrt[2]{2\frac{800me}{100me}}=\frac{2\sqrt[2]{22}}{1}$