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Question 73 of 83

Q.The kinetic energy of an alpha particle is four times the kinetic energy of a proton. The ratio λαλp\dfrac{\lambda_{\alpha}}{\lambda_{p}} of the de Broglie wavelengths associated with them will be: (A) 116\dfrac{1}{16} (B) 18\dfrac{1}{8} (C) 14\dfrac{1}{4} (D) 12\dfrac{1}{2}

Assam AhsecCBSE Class XII Board 2025MCQ· 1mImportance★★★★★
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The de Broglie wavelength depends on momentum, not directly on kinetic energy. Using K=p22mK = \frac{p^2}{2m} and the given Kα=4KpK_\alpha = 4K_p, along with mα=4mpm_\alpha = 4m_p, we find λαλp=14\frac{\lambda_\alpha}{\lambda_p} = \frac{1}{4}, which corresponds to option (C).

The de Broglie wavelength is the bridge between particle and wave behaviour: λ=hp\lambda = \frac{h}{p}, where hh is Planck’s constant and pp is the momentum. The problem gives you kinetic energy, not momentum directly — so the first step is always to connect KK and pp.

For any non-relativistic particle, kinetic energy is K=p22mK = \frac{p^2}{2m}. Rearranging, p=2mKp = \sqrt{2mK}. This is the key relation that lets you translate the given energy ratio into a wavelength ratio.

  1. Write the wavelength for each particle.

    For the alpha particle: λα=hpα=h2mαKα\lambda_\alpha = \frac{h}{p_\alpha} = \frac{h}{\sqrt{2 m_\alpha K_\alpha}}.

    For the proton: λp=hpp=h2mpKp\lambda_p = \frac{h}{p_p} = \frac{h}{\sqrt{2 m_p K_p}}.

  2. Take the ratio.

λαλp=h/2mαKαh/2mpKp=mpKpmαKα.\frac{\lambda_\alpha}{\lambda_p} = \frac{h / \sqrt{2 m_\alpha K_\alpha}}{h / \sqrt{2 m_p K_p}} = \sqrt{\frac{m_p K_p}{m_\alpha K_\alpha}}.

Notice that hh and the factor 22 cancel out neatly.

  1. Plug in the given data.

    You are told Kα=4KpK_\alpha = 4 K_p. Also, an alpha particle is a helium nucleus — 2 protons and 2 neutrons — so its mass is approximately 4 times the proton mass: mα=4mpm_\alpha = 4 m_p.

    Substitute:

    λαλp=mp⋅Kp(4mp)⋅(4Kp)=116=14.\frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p \cdot K_p}{(4 m_p) \cdot (4 K_p)}} = \sqrt{\frac{1}{16}} = \frac{1}{4}. …

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