
The momentum of a particle having a de- Broglie wavelength of ${10^{ - 17}}{\text{m}}$ is:
(Given: ${\text{h}} = 6.625 \times {10^{ - 34}}{\text{m}}$)
A. $3.3125 \times {10^{ - 7}}{\text{kg m }}{{\text{S}}^{ - 1}}$
B. $26.5 \times {10^{ - 7}}{\text{kg m }}{{\text{S}}^{ - 1}}$
C. $6.625 \times {10^{ - 17}}{\text{kg m }}{{\text{S}}^{ - 1}}$
D. $13.25 \times {10^{ - 17}}{\text{kg m }}{{\text{S}}^{ - 1}}$
Answer
569.4k+ views
Hint: To answer this question, you must recall the formula for de Broglie’s wavelength of an electron. De Broglie proposed a theory suggesting that every form of matter behaves like waves in some or the other circumstances.
Formula used:
$\lambda = \dfrac{h}{{mv}} = \dfrac{h}{p}$
Where, $\lambda $ is the de- Broglie wavelength of the matter wave
$h$ is Planck’s constant
$m$ is the mass of the given particle under consideration
$v$ is the velocity of the given particle under consideration
And $p$ is the momentum of the particle
Complete step by step answer:
We are supposed to find the momentum of a particle whose wavelength is provided to us. We can use the de- Broglie equation directly to get the momentum of the particle.
It is given to us in the question that the value of Planck’s constant is given as ${\text{h}} = 6.625 \times {10^{ - 34}}{\text{m}}$.
We know from the de- Broglie equation that $\lambda = \dfrac{h}{{mv}} = \dfrac{h}{p}$.
Or we can write, $p = \dfrac{h}{\lambda }$.
Substituting the values, we get,
$p = \dfrac{{6.625 \times {{10}^{ - 34}}}}{{{{10}^{ - 17}}}}$
$\therefore p = 6.625 \times {10^{ - 17}}{\text{ kg m }}{{\text{s}}^{ - 1}}$
Thus, the correct answer is C.
Note:
Matter waves are a crucial part of the quantum mechanical theory, being an example of the dual nature of matter. It was suggested that all matter particles exhibit a wave-like behaviour which may or may not be significant enough. For instance, a beam of electrons is diffracted in the same way like a beam of light or a water wave does. In most cases, the wavelength of objects is too small to have a significant impact on our day-to-day activities. Hence in our day-to-day lives, with objects of the size of tennis balls or with people, matter waves are not of significant wavelength. These matter waves are referred to as de Broglie waves.
Formula used:
$\lambda = \dfrac{h}{{mv}} = \dfrac{h}{p}$
Where, $\lambda $ is the de- Broglie wavelength of the matter wave
$h$ is Planck’s constant
$m$ is the mass of the given particle under consideration
$v$ is the velocity of the given particle under consideration
And $p$ is the momentum of the particle
Complete step by step answer:
We are supposed to find the momentum of a particle whose wavelength is provided to us. We can use the de- Broglie equation directly to get the momentum of the particle.
It is given to us in the question that the value of Planck’s constant is given as ${\text{h}} = 6.625 \times {10^{ - 34}}{\text{m}}$.
We know from the de- Broglie equation that $\lambda = \dfrac{h}{{mv}} = \dfrac{h}{p}$.
Or we can write, $p = \dfrac{h}{\lambda }$.
Substituting the values, we get,
$p = \dfrac{{6.625 \times {{10}^{ - 34}}}}{{{{10}^{ - 17}}}}$
$\therefore p = 6.625 \times {10^{ - 17}}{\text{ kg m }}{{\text{s}}^{ - 1}}$
Thus, the correct answer is C.
Note:
Matter waves are a crucial part of the quantum mechanical theory, being an example of the dual nature of matter. It was suggested that all matter particles exhibit a wave-like behaviour which may or may not be significant enough. For instance, a beam of electrons is diffracted in the same way like a beam of light or a water wave does. In most cases, the wavelength of objects is too small to have a significant impact on our day-to-day activities. Hence in our day-to-day lives, with objects of the size of tennis balls or with people, matter waves are not of significant wavelength. These matter waves are referred to as de Broglie waves.
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