Question

A free particle with initial kinetic energy E and de-Broglie wavelength λ enters a region in which it has potential energy U. What is the particle’s new de-Broglie wavelength?
A. $\lambda {\left( {1 - \dfrac{U}{E}} \right)^{ - \dfrac{1}{2}}}$
B. $\lambda \left( {1 - \dfrac{U}{E}} \right)$
C. $\lambda {\left( {1 - \dfrac{U}{E}} \right)^{ - 1}}$
D. $\lambda {\left( {1 - \dfrac{U}{E}} \right)^{\dfrac{1}{2}}}$

Answer 1

De-Broglie particle wavelength of a particle is given by the formula $\lambda = \dfrac{h}{{\sqrt {2mE} }}$, where m is the mass of the particle, and E is the energy of the particle. A free particle is a particle that is not bound by any external forces, and its potential energy is constant.
The de-Broglie wavelength of a particle indicates the length at which a wave repeats its property for a particle. De-Broglie is represented by a symbol λ, and for a particle with momentum p, the de-Broglie wavelength is given by the formula $\lambda = \dfrac{h}{p}$.
Here, in this question, we need to determine the new de-Broglie wavelength of the particle such that the particle enters into a region having potential energy U and the initial kinetic energy K.

Complete step by step answer:
A free particle is characterized by a fixed velocity v, and its momentum is given by $p = mv$ since kinetic energy of a particle is given by $E = \dfrac{1}{2}m{v^2}$, so the total energy will be equal to
$E = \dfrac{1}{2}m{v^2} = \dfrac{{{p^2}}}{{2m}}$
Hence initial kinetic energy will be

$$E = \dfrac{{{p^2}}}{{2m}} \\\ p = \sqrt {2mE} \\\$$

The de-Broglie wavelength of a particle is given as
$\lambda = \dfrac{h}{p}$
Also, we can write
$\lambda = \dfrac{h}{{\sqrt {2mE} }}$
Since a particle enters a region in which the potential energy is U, hence the energy of the particle when it enters the region will be
${E_f} = E - U$
Hence the wavelength in the region becomes
${\lambda _f} = \dfrac{h}{{\sqrt {2m{E_f}} }}$
${\lambda _f} = \dfrac{h}{{\sqrt {2m\left( {E - U} \right)} }}$ [Since${E_f} = E - U$]
${\lambda _f} = \dfrac{h}{{\sqrt {2m\left( {E - U} \right)} }}$
This can be written as
${\lambda _f}^2 = \dfrac{{{h^2}}}{{2m\left( {E - U} \right)}}$
Now take E as common, we get

$${\lambda _f}^2 = \dfrac{{{h^2}}}{{2m\left( {E - U} \right)}} \\\ {\lambda _f}^2 = \dfrac{{{h^2}}}{{2mE\left( {1 - \dfrac{U}{E}} \right)}} \\\$$

We can write as

$${\lambda _f} = \dfrac{h}{{\sqrt {2mE\left( {1 - \dfrac{U}{E}} \right)} }} \\\ {\lambda _f} = \dfrac{h}{{\sqrt {2mE} }} \times \dfrac{1}{{{{\left( {1 - \dfrac{U}{E}} \right)}^{\dfrac{1}{2}}}}} \\\$$

Since$\lambda = \dfrac{h}{{\sqrt {2mE} }}$hence we can write
${\lambda _f} = \lambda \times {\left( {1 - \dfrac{U}{E}} \right)^{ - \dfrac{1}{2}}}$
Hence the particle’s new de-Broglie wavelength is$= \lambda \times {\left( {1 - \dfrac{U}{E}} \right)^{ - \dfrac{1}{2}}}$
Option A is correct.

Note: Students should always keep in mind that the energy is conserved in nature, and so as here, the kinetic energy of the particle got transversed to the potential energy.