
The figure shows a small particle in a box which bounces back and forth at constant speed. The reflections of particles from the wall create a standing wave analogous to standing waves on a string tied at both ends. Since a standing wave confined to a region can only have selected wavelengths, momentum of the particle is quantized. We can safely assume that such a particle only has kinetic energy. This energy must also be quantized. [There may be multiple correct answers.]
(A) The momentum of particle in nth node of standing wave is
(B) Particle’s energy in nth state is
(C) Matter waves in this case will be electromagnetic in nature.
(D) None of these.
Answer
561k+ views
Hint:
De-Broglie’s hypotheses are used when we have a particle behaving like a wave. Using the quantization equations, the momentum and energy of the particle at nodes of the wave can be found.
Formula used: $\lambda = \dfrac{h}{p}$ gives the wavelength of the particle with momentum p. h is the Planck’s constant.
Complete step by step answer:
The equations given by de Broglie describe the wavelength of the object given its mass and velocity with which it travels.
In this question, we have a particle behaving as a wave. As it travels back and forth, the wavelength of the particle will be given as:
$\lambda = 2L$
For nth node this wavelength will be given as:
${\lambda _n} = \dfrac{\lambda }{n} = \dfrac{{2L}}{n}$
By de Broglie’s equation, we know that:
${\lambda _n} = \dfrac{h}{p} = \dfrac{{2L}}{n}$
Solving for the momentum p, we get:
$p = \dfrac{{nh}}{{2L}}$
Now, moving on to the energy of a particle. It is given as:
${E_n} = \dfrac{{{p^2}}}{{2m}} = \dfrac{{{{\left( {\dfrac{{nh}}{{2L}}} \right)}^2}}}{{2m}}$
${E_n} = \dfrac{{{n^2}{h^2}}}{{8m{L^2}}}$
We already know that matter waves are not electromagnetic in nature, as EM waves are only produced by charged particles.
Hence, the correct answers are option (A) and option (C).
Note:
It was proposed by de Broglie that light exhibits both wave and particle-like properties and so does matter. This concept played a vital role in the invention of the electron microscope for observation of atomic activities at the minuscule level.
De-Broglie’s hypotheses are used when we have a particle behaving like a wave. Using the quantization equations, the momentum and energy of the particle at nodes of the wave can be found.
Formula used: $\lambda = \dfrac{h}{p}$ gives the wavelength of the particle with momentum p. h is the Planck’s constant.
Complete step by step answer:
The equations given by de Broglie describe the wavelength of the object given its mass and velocity with which it travels.
In this question, we have a particle behaving as a wave. As it travels back and forth, the wavelength of the particle will be given as:
$\lambda = 2L$
For nth node this wavelength will be given as:
${\lambda _n} = \dfrac{\lambda }{n} = \dfrac{{2L}}{n}$
By de Broglie’s equation, we know that:
${\lambda _n} = \dfrac{h}{p} = \dfrac{{2L}}{n}$
Solving for the momentum p, we get:
$p = \dfrac{{nh}}{{2L}}$
Now, moving on to the energy of a particle. It is given as:
${E_n} = \dfrac{{{p^2}}}{{2m}} = \dfrac{{{{\left( {\dfrac{{nh}}{{2L}}} \right)}^2}}}{{2m}}$
${E_n} = \dfrac{{{n^2}{h^2}}}{{8m{L^2}}}$
We already know that matter waves are not electromagnetic in nature, as EM waves are only produced by charged particles.
Hence, the correct answers are option (A) and option (C).
Note:
It was proposed by de Broglie that light exhibits both wave and particle-like properties and so does matter. This concept played a vital role in the invention of the electron microscope for observation of atomic activities at the minuscule level.
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