Given Data:

• Mass of proton ($m_p$) = $m$
• Charge of proton ($q_p$) = $e$
• Mass of alpha particle ($m_\alpha$) = $4m$
• Charge of alpha particle ($q_\alpha$) = $2e$

Step 1 (Formula / Governing Law): [0.5 Mark]

The de Broglie wavelength ($\lambda$) is given by the relation:

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

Where $h$ is Planck's constant and $p$ is the momentum of the particle.

Step 2 (Step-by-Step Mathematical Working): [1 Mark]

(a) Same speeds ($v$):

Since $p = mv$, then $\lambda = \frac{h}{mv}$.

$$\frac{\lambda_\alpha}{\lambda_p} = \frac{m_p v}{m_\alpha v} = \frac{m}{4m} = \frac{1}{4}$$

(b) Same kinetic energy ($K$):

Since $p = \sqrt{2mK}$, then $\lambda = \frac{h}{\sqrt{2mK}}$.

$$\frac{\lambda_\alpha}{\lambda_p} = \frac{\sqrt{2m_p K}}{\sqrt{2m_\alpha K}} = \sqrt{\frac{m}{4m}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$$

(c) Same potential difference ($V$):

Since $K = qV$, then $p = \sqrt{2mqV}$, so $\lambda = \frac{h}{\sqrt{2mqV}}$.

$$\frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p q_p}{m_\alpha q_\alpha}} = \sqrt{\frac{m \cdot e}{4m \cdot 2e}} = \sqrt{\frac{1}{8}} = \frac{1}{2\sqrt{2}}$$

Step 3 (Final Concluded Answer): [0.5 Mark]

The ratios of the de Broglie wavelengths are:

• (a) 1 : 4
• (b) 1 : 2
• (c) 1 : 2\sqrt{2}

💡 EL10 Concept Insight:

Think of the de Broglie wavelength as a "wave-like" shadow cast by a moving particle; the heavier or faster the particle is, the "shorter" and more compressed its wave becomes, which is why the alpha particle (being 4 times heavier) always has a shorter wavelength than the proton in these scenarios.