Question

A proton, a neutron, an electron and an α-particle have same energy. Then their de Broglie wavelengths compare as ______.

पर्याय

• λ_p = λ_n > λ_e > λ_α

• λ_α < λ_p = λ_n > λ_e

• λ_e < λ_p = λ_n > λ_α

• λ_e = λ_p = λ_n = λ_α

Answer 1

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, m_p = m_n, hence λ_p = λ_n

As m_α > m_p therefore λ_α < λ_p

As m_e > m_n therefore λ_e < λ_n

Hence. λ_α < λ_p = λ_n < λ_e