Question

A proton and an alpha - particle are accelerated through same potential difference. Then, the ratio of de-Broglie wavelength of proton and alpha-particle is:

• A. $\sqrt{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $2\sqrt{2}$
• D. None of these

Answer 1

The correct option is C

$2\sqrt{2}$

Step 1: Given

Potential difference=$V$

Step 2: Determine the de-Broglie wavelength of a proton

The de-Broglie wavelength of a particle is given by $\lambda =\frac{h}{P}$ (where $h$ is the Planck's constant and $P$ is the momentum)

Let the mass of the proton be $m_p$ and the momentum of the proton is given by $p=\sqrt{2m_{p}k}$ (where $k$ is the kinetic energy)

The de-Broglie wavelength will be ${\lambda }_p=\frac{h}{\sqrt{2m_{p}k}}$

Kinetic energy for a proton in an electric field can be rewritten as $k=q_{p}V$

Therefore, ${\lambda }_p=\frac{h}{\sqrt{2m_p{q}_{p}V}}\left(1\right)$

Step 3: Determine the de-Broglie wavelength of an alpha particle

Let the mass of an alpha particle be $m_a$

Similarly, the de-Broglie wavelength will be ${\lambda }_a=\frac{h}{\sqrt{2m_a{q}_{a}V}}\left(2\right)$

Step 4: Compare the properties of a proton and an alpha-particle

$q_{a}and{q}_p$ are the charges on the alpha particle and the proton.

Comparing the charges, $q_a=2q_p$

Comparing the masses, $m_a=4m_p$

Step 5: Substituting the above comparisons in equation $\left(2\right)$

$$\begin{gathered} {\lambda }_a=\frac{h}{\sqrt{2m_a{q}_{a}V}} \\ =\frac{h}{\sqrt{2\times 4m_a\times 2q_p\times V}}\left(3\right) \end{gathered}$$

Step 6: Determine the ratio of equations $\left(1\right)$ and $\left(3\right)$

$$\begin{gathered} \frac{{\lambda }_p}{{\lambda }_a}=\frac{\frac{h}{\sqrt{2m_p{q}_{p}V}}}{\frac{h}{\sqrt{2\times 4m_a\times 2q_p\times V}}} \\ =\frac{4}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}} \\ =2\sqrt{2} \end{gathered}$$

Hence, option C is the correct option.