
The wavelength associated with the electron motion
A. Increases with increase in the speed of the electron
B. Remains the same irrespective of the speed of the electron
C. Decreases with the increase in speed of the electron
D. Is zero
Answer
479.7k+ views
Hint: To know the answer to this question, we need to know about the de Broglie equation. We come to know from the de Broglie equation that wavelength is inversely proportional to the velocity of the particle. Therefore wavelength associated with the electron motion decreases with an increase in the speed of the electron.
Formula Used:
\[\lambda = \dfrac{h}{p}\]
This formula is the famous de Broglie equation.
Here, \[h\]is Planck’s constant
\[p\] is the linear momentum
\[\lambda \] is the wavelength
Complete step by step solution:
From the above formula used, we can substitute \[p = mv\]
So we get,
\[\lambda = \dfrac{h}{{mv}}\]
From the above expression, we can conclude that the wavelength associated with any particle is inversely proportional to the velocity of that particle.
So we can conclude that the wavelength associated with the electron motion decreases with an increase in the speed of the electron.
Therefore the correct option is C.
Additional Information:
In \[1924\] , Louis de Broglie introduced his theory, in which he proposed electrons have properties of the two waves and particles, similar to light. He improved the details of the Plank-Einstein connection to apply to a wide range of issues. According to de Broglie, every moving particle sometimes acts as a particle and sometimes as a wave. We call the wave associated with the moving particle as a matter-wave or de Broglie wave. And the wavelength associated with this wave is called the de Broglie wavelength.
Note:
Note that a ball and a car are also moving particles. We cannot say that a ball and a car are waves. Later de Broglie calculated the wavelength of a cricket ball and found that it is immeasurable. So they concluded that de Broglie’s theory of wave-particle duality is valid for the moving objects which are up to the size of the electrons.
Formula Used:
\[\lambda = \dfrac{h}{p}\]
This formula is the famous de Broglie equation.
Here, \[h\]is Planck’s constant
\[p\] is the linear momentum
\[\lambda \] is the wavelength
Complete step by step solution:
From the above formula used, we can substitute \[p = mv\]
So we get,
\[\lambda = \dfrac{h}{{mv}}\]
From the above expression, we can conclude that the wavelength associated with any particle is inversely proportional to the velocity of that particle.
So we can conclude that the wavelength associated with the electron motion decreases with an increase in the speed of the electron.
Therefore the correct option is C.
Additional Information:
In \[1924\] , Louis de Broglie introduced his theory, in which he proposed electrons have properties of the two waves and particles, similar to light. He improved the details of the Plank-Einstein connection to apply to a wide range of issues. According to de Broglie, every moving particle sometimes acts as a particle and sometimes as a wave. We call the wave associated with the moving particle as a matter-wave or de Broglie wave. And the wavelength associated with this wave is called the de Broglie wavelength.
Note:
Note that a ball and a car are also moving particles. We cannot say that a ball and a car are waves. Later de Broglie calculated the wavelength of a cricket ball and found that it is immeasurable. So they concluded that de Broglie’s theory of wave-particle duality is valid for the moving objects which are up to the size of the electrons.
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