CBSE 2025, Set 1

Question

The kinetic energy of an alpha particle is four times the kinetic energy of a proton. The ratio of de Broglie wavelengths $\left(\dfrac{\lambda_\alpha}{\lambda_p}\right)$ associated with them will be : A). $\dfrac{1}{16}$ B). $\dfrac{1}{8}$ C). $\dfrac{1}{4}$ D). $\dfrac{1}{2}$

Edvaya · Accepted Answer

The correct answer is $\dfrac{1}{4}$. The de Broglie wavelength is: $$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$$ where $K$ is the kinetic energy. Given: $K_\alpha = 4K_p$ Mass of alpha particle: $m_\alpha = 4m_p$ (since it contains 2 protons + 2 neutrons) $$\frac{\lambda_\alpha}{\lambda_p} = \frac{h/\sqrt{2m_\alpha K_\alpha}}{h/\sqrt{2m_p K_p}} = \sqrt{\frac{m_p K_p}{m_\alpha K_\alpha}}$$ $$= \sqrt{\frac{m_p \cdot K_p}{4m_p \cdot 4K_p}} = \sqrt{\frac{1}{16}} = \boxed{\frac{1}{4}}$$ > **Step-by-step check:** $m_\alpha = 4m_p$ and $K_\alpha = 4K_p$, so the product $m_\alpha K_\alpha = 16 m_p K_p$. The ratio of wavelengths is therefore $1/\sqrt{16} = 1/4$.

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Physics > Modern Physics > Dual Nature Of Radiation And Matter

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