Question

The potential energy of a particle varies as

\(U\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {E_0}& {0\le x \le 1}\\ {0}&{x > 1} \end{array}} \right.\)

For 0 ≤ x ≤ 1. De-Broglie wavelength is λ₁ and for x > 1 the de-Broglie wavelength is λ₂. The total energy of the particle is 2E₀. Find \(\frac{{{\lambda _1}}}{{{\lambda _2}}}\)

1. 1
2. √2
3. 2√2
4. 2

Answer 1

Correct Answer - Option 2 : √2

Concept:

de Broglie wavelength of electrons: Louis de Broglie theorized that not only light possesses both wave and particle properties, but rather particles with mass - such as electrons - do as well.

The wavelength of material waves is also known as the de Broglie wavelength.

de Broglie wavelength (λ) of electrons can be calculated from Planks constant h divided by the momentum of the particle.

λ = h/p

where λ is de Broglie wavelength, h is Plank's const, and p is the momentum.

In terms of Energy, de Broglie wavelength of electrons:

\(λ=\frac{h}{\sqrt{2mE}}\)

where λ is de Broglie wavelength, h is Plank's const, m is the mass of an electron, and E is the energy of the electron.

Calculation:

For 0 ≤ x ≤ 1, PE = E₀

∴ Kinetic energy K₁ = Total energy – PE

= 2E₀ – E₀ = E₀

∴ \({\lambda _1} = \frac{h}{{\sqrt {2m{E_0}} }}\)     …i)

For x > 1, PE = 0

∴ Kinetic energy K₂ = Total energy = 2E₀

∴ \({\lambda _2} = \frac{h}{{\sqrt {4m{E_0}} }}\)     …ii)

From Eqs. i) and ii), we have

\(\frac{{{\lambda _1}}}{{{\lambda _2}}} = \sqrt 2 \)