Matter Waves Energy Relation Shortcut

 

❓ Question

Find the ratio of the de Broglie wavelengths of:

• a deuteron having kinetic energy $E$,
• an $\alpha$-particle having kinetic energy $2E$.

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✍️ Short Explanation

For a non-relativistic particle, the de Broglie wavelength is

$$\lambda=\frac{h}{\sqrt{2mK}},$$

where

• $h$ = Planck's constant,
• $m$ = mass of the particle,
• $K$ = kinetic energy.

Thus,

$$\lambda \propto \frac{1}{\sqrt{mK}}.$$

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🔷 Step 1 — Write Masses and Energies

Deuteron

• Mass:

$$m_d \approx 2m$$

• Kinetic energy:

$$K_d=E.$$

$\alpha$-particle

• Mass:

$$m_\alpha \approx 4m$$

• Kinetic energy:

$$K_\alpha=2E.$$

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🔷 Step 2 — Find the Ratio

Since

$$\lambda \propto \frac{1}{\sqrt{mK}},$$
$$\frac{\lambda_d}{\lambda_\alpha} = \sqrt{\frac{m_\alpha K_\alpha}{m_d K_d}}.$$

Substituting the values,

$$\frac{\lambda_d}{\lambda_\alpha} = \sqrt{\frac{(4m)(2E)}{(2m)(E)}} = \sqrt{\frac{8mE}{2mE}} = \sqrt{4} = 2.$$

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✅ Final Answer

$$\boxed{\lambda_d:\lambda_\alpha = 2:1}$$

or equivalently,

$$\boxed{\frac{\lambda_d}{\lambda_\alpha}=2.}$$

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🔷 JEE Shortcut

Use the direct relation:

$$\boxed{\lambda \propto \frac{1}{\sqrt{mK}}}$$

Then,

$$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2K_2}{m_1K_1}}.$$

Here,

$$\sqrt{\frac{4\times 2}{2\times 1}} = \sqrt{4} = \boxed{2}.$$

So, the required ratio is

$$\boxed{2:1}.$$

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