Q.Find the ratio of the de Broglie wavelengths and associated respectively with an alpha particle and a proton,
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Start your 14-day free trial to unlock the full solution →The de Broglie wavelength depends on momentum, which relates to kinetic energy through mass. For equal kinetic energy, the heavier alpha particle has larger momentum and shorter wavelength: . For equal accelerating voltage, the alpha's greater charge means it gains more energy, giving an even smaller ratio: .
The de Broglie wavelength connects the wave and particle nature of matter through , where momentum 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 and , respectively. An alpha particle (helium nucleus) has mass and charge .
(i) Same Kinetic Energy
When two particles have the same kinetic energy , we need to compare their momenta.
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Relate momentum to kinetic energy. For a non-relativistic particle, kinetic energy is , which gives momentum .
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Express the wavelength ratio. Using de Broglie's relation:
- Substitute the momentum expressions. Since both have the same kinetic energy :
- Use the mass ratio. With :
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 , the energy gained depends on their charge.
- Find the kinetic energy gained. A particle with charge accelerated through potential difference gains kinetic energy:
So the proton gains while the alpha particle gains .
- Express momenta in terms of voltage. Using : …
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