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If the momentum of an electron is changed by P, then the de-Broglie wavelength associated with it changes by 0.5%. The initial momentum of an electron will be:A. 400 PB. $\dfrac{\text{P}}{200}$C. 100 PD. 200 P

Last updated date: 06th Sep 2024
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Hint: In this problem, we use the equation which was given by de-Broglie. He gave the relationship between the momentum of the particle, the Planck's constant and wavelength i.e. $\lambda \text{ = }\dfrac{\text{h}}{\text{P}}$. From here, we can calculate the value of initial momentum.

Complete step by step Answer:
- In the given question, we have to calculate the initial momentum of an electron from the given data.
- According to the de-Broglie, the wavelength of an object that is related to the momentum and mass of the object is known as de-Broglie wavelength.
- Now, as we know that the relationship is given by:
$\lambda \text{ = }\dfrac{\text{h}}{\text{P}}$ or $\text{P = }\dfrac{\text{h}}{\lambda }$ …. (1)
- Now, it is given that the final momentum of an electron is 0.5%, so we can write the equation (1) as
$\dfrac{\vartriangle \text{P}}{\text{P}}\ \text{= - }\dfrac{\vartriangle \lambda }{\lambda }$
- The negative sign signifies that the change in the momentum will be opposite to the change in the wavelength.
- So, here we have to the find the value of P which is initial momentum so by putting the value of wavelength we will get:
$\dfrac{\vartriangle \text{P}}{\text{P}}\ \text{= }\dfrac{0.5}{100}$
$\text{P = }\dfrac{100}{0.5}\vartriangle \text{P = 200}\vartriangle \text{P}$
- So, we can say that the initial momentum is equal to the 200 times of the final momentum.

Therefore, option D is the correct answer.

Note: According to de-Broglie the matter consists of dual nature that is particle and wave nature just like the light. We can study the properties of the matter waves of the very small objects. In de-Broglie wavelength, the momentum is defined as the product of the mass and velocity of the object.