According to de Broglie, the relation between the wavelength of the wave associated with the moving electron and momentum of the electron is $\lambda = \dfrac{h}{{mv}}$
A.True
B.False
Answer
601.5k+ views
Hint: To answer this question, you must recall the de Broglie’s equation. The de Broglie equation states that all matter acts as particles, and as waves, like light and radiation. This equation gives us the wave nature of all moving particles whether it is macroscopic or microscopic.
Complete answer:
The de Broglie equation denotes the wave nature of an electron.
It was seen earlier that electromagnetic radiations exhibit dual nature, i.e., they behave both like a particle (having momentum) and like a wave (expressed in frequency, wavelength). This dual nature was first studied in the photoelectric effect.
de Broglie proposed that any moving particle, whether it be microscopic or macroscopic is bound to be associated with a wave character. These waves were termed as matter waves. He also gave an expression between the momentum of a particle with its wavelength thus, giving a relation between the particle and wave nature.
From the Planck’s quantum law, it is known that energy of an electromagnetic wave can be given by $E = h\nu = h\dfrac{c}{\lambda }$
Also, Einstein gave the energy of a particle of matter to be equal to $E = m{c^2}$or $E = m{v^2}$
Since de Broglie suggested that matter has both wave and particle nature, hence these two equations must be equal. We get,
$E = h\dfrac{c}{\lambda } = m{v^2}$
Thus, from this equation, we can say that, $\lambda = \dfrac{h}{{mv}}$
This equation is the De Broglie equation. It gives us the relation between the wavelength of the wave associated with the movement of the electron and momentum of the electron as a particle.
Thus, the given statement is true.
Hence, the correct answer is A.
Note:
The experimental verification of the de Broglie equation was first given by the Davisson Germer experiment in which they recorded the pattern observed by scattering electrons at fixed energy values and observed a pattern similar to that of X- rays.
Complete answer:
The de Broglie equation denotes the wave nature of an electron.
It was seen earlier that electromagnetic radiations exhibit dual nature, i.e., they behave both like a particle (having momentum) and like a wave (expressed in frequency, wavelength). This dual nature was first studied in the photoelectric effect.
de Broglie proposed that any moving particle, whether it be microscopic or macroscopic is bound to be associated with a wave character. These waves were termed as matter waves. He also gave an expression between the momentum of a particle with its wavelength thus, giving a relation between the particle and wave nature.
From the Planck’s quantum law, it is known that energy of an electromagnetic wave can be given by $E = h\nu = h\dfrac{c}{\lambda }$
Also, Einstein gave the energy of a particle of matter to be equal to $E = m{c^2}$or $E = m{v^2}$
Since de Broglie suggested that matter has both wave and particle nature, hence these two equations must be equal. We get,
$E = h\dfrac{c}{\lambda } = m{v^2}$
Thus, from this equation, we can say that, $\lambda = \dfrac{h}{{mv}}$
This equation is the De Broglie equation. It gives us the relation between the wavelength of the wave associated with the movement of the electron and momentum of the electron as a particle.
Thus, the given statement is true.
Hence, the correct answer is A.
Note:
The experimental verification of the de Broglie equation was first given by the Davisson Germer experiment in which they recorded the pattern observed by scattering electrons at fixed energy values and observed a pattern similar to that of X- rays.
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