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

Hint: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.