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प्रश्न
A proton, a neutron, an electron and an α-particle have same energy. Then their de Broglie wavelengths compare as ______.
पर्याय
λp = λn > λe > λα
λα < λp = λn > λe
λe < λp = λn > λα
λe = λp = λn = λα
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उत्तर
A proton, a neutron, an electron and an α-particle have same energy. Then their de Broglie wavelengths compare as `underline(λ_α < λ_p = λ_n > λ_e)`.
Explanation:
According to de-Broglie a moving material particle sometimes acts as a wave and sometimes as a particle.
deiBroglie wavelength: According to de-Broglie theory, the wavelength of de-Broglie wave is given by
`λ = h/p = h/(mv) = h/sqrt(2mE) ⇒ λ oo 1/p oo 1/v oo 1/sqrt(E)`
Where h = Plank's constant, m = Mass of the particle, v = Speed of the particle, E = Energy of the particle.
The smallest wavelength whose measurement is possible is that of γ-rays.
The wavelength of matter waves associated with the microscopic particles like electron, proton neutron, α-particle etc is of the order of 10–10 m.
We conclude from above that the relation between de-Broglie wavelength λ and kinetic energy K of the particle is given by
`λ = h/sqrt(2mK)`
Here, for the given value of energy K, `h/sqrt(2K)` is a constant.
Thus, `λ oo 1/sqrt(m)`
∴ `λ_p : λ_n : λ_e : λ_α = 1/sqrt(m_p) : 1/sqrt(m_n) : 1/sqrt(m_e) : 1/sqrt(m_α)`
Comparing the wavelength of proton and neutron, mp = mn, hence λp = λn
As mα > mp therefore λα < λp
As me > mn therefore λe < λn
Hence. λα < λp = λn < λe
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