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Question 75 of 83

Q.Find the ratio (λaλp)\left(\dfrac{\lambda_a}{\lambda_p}\right) of the de Broglie wavelengths λa\lambda_a and λp\lambda_p associated respectively with an alpha particle and a proton,

(i) if they are moving with the same kinetic energy;
(ii) just after they are accelerated through the same potential difference.
Odisha ChseCBSE Class XII Board 2026Subjective· 2mImportance★★★★★
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The de Broglie wavelength λ=hp\lambda = \frac{h}{p} depends on momentum, which relates to kinetic energy through mass. For equal kinetic energy, the heavier alpha particle has larger momentum and shorter wavelength: λaλp=12\frac{\lambda_a}{\lambda_p} = \frac{1}{2}. For equal accelerating voltage, the alpha's greater charge means it gains more energy, giving an even smaller ratio: λaλp=122\frac{\lambda_a}{\lambda_p} = \frac{1}{2\sqrt{2}}.

The de Broglie wavelength connects the wave and particle nature of matter through λ=hp\lambda = \frac{h}{p}, where momentum pp is the bridge. The key insight is that while wavelength depends inversely on momentum, momentum itself relates to kinetic energy differently depending on the particle's mass. A heavier particle carrying the same kinetic energy must be moving slower but has greater momentum—and therefore a shorter wavelength.

Let's denote the mass and charge of a proton as mpm_p and ee, respectively. An alpha particle (helium nucleus) has mass ma=4mpm_a = 4m_p and charge qa=2eq_a = 2e.

(i) Same Kinetic Energy

When two particles have the same kinetic energy KK, we need to compare their momenta.

  1. Relate momentum to kinetic energy. For a non-relativistic particle, kinetic energy is K=p22mK = \frac{p^2}{2m}, which gives momentum p=2mKp = \sqrt{2mK}.

  2. Express the wavelength ratio. Using de Broglie's relation:

λaλp=h/pah/pp=pppa\frac{\lambda_a}{\lambda_p} = \frac{h/p_a}{h/p_p} = \frac{p_p}{p_a}

  1. Substitute the momentum expressions. Since both have the same kinetic energy KK:

λaλp=2mpK2maK=mpma\frac{\lambda_a}{\lambda_p} = \frac{\sqrt{2m_p K}}{\sqrt{2m_a K}} = \sqrt{\frac{m_p}{m_a}}

  1. Use the mass ratio. With ma=4mpm_a = 4m_p:

λaλp=mp4mp=14=12\frac{\lambda_a}{\lambda_p} = \sqrt{\frac{m_p}{4m_p}} = \sqrt{\frac{1}{4}} = \frac{1}{2}

Tip

For particles with the same kinetic energy, the wavelength ratio equals the square root of the inverse mass ratio: lighter particles have longer wavelengths.

(ii) Same Accelerating Potential Difference

When particles are accelerated through the same potential difference VV, the energy gained depends on their charge.

  1. Find the kinetic energy gained. A particle with charge qq accelerated through potential difference VV gains kinetic energy:

K=qVK = qV

So the proton gains Kp=eVK_p = eV while the alpha particle gains Ka=2eVK_a = 2eV.

  1. Express momenta in terms of voltage. Using p=2mKp = \sqrt{2mK}: …

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