The de-Broglie wavelength associated with proton changes by 0.25% if its momentum is changed by P₀. The initial momentum wasa) 100 P₀ b) P₀/400 c) 401 P₀ d) P₀/100 Correct answer is option 'C'. Can you explain this answer?

Question

The de-Broglie wavelength associated with proton changes by 0.25% if its momentum is changed by P₀. The initial momentum wasa)                                  100 P₀                               b)                                  P₀/400                               c)                                  401 P₀                               d)                                  P₀/100                               Correct answer is option 'C'. Can you explain this answer?

Answer

L= wavelength. -(∆L/L)X100=(∆P/P)X 100. 0.25=(Po/P)X100. P=Po(100X100)/25=400Po.

Answer

Use the formula lamba=h/p you will see that they are inversely related, so if you reduce wavelength by 0.25% then momentum reduces by 1 unit. So that implies Initial momentum is 401Po

Answer

Concept:

The de-Broglie wavelength associated with a particle is given by the equation:

$$\lambda = \frac{h}{p}$$

where $\lambda$ is the de-Broglie wavelength, $h$ is the Planck's constant, and $p$ is the momentum of the particle.

Solution:

To find the initial momentum of the proton, we can use the given information that the de-Broglie wavelength changes by 0.25% when the momentum changes by $P_0$.

Let's denote the initial momentum of the proton as $P_i$ and the final momentum as $P_f$.

Given that:

$$\frac{\lambda_f - \lambda_i}{\lambda_i} = \frac{0.25}{100} = 0.0025$$

We know that:

$$\lambda_i = \frac{h}{P_i}$$

and

$$\lambda_f = \frac{h}{P_f}$$

Substitute these into the percentage change formula:

$$\frac{\frac{h}{P_f} - \frac{h}{P_i}}{\frac{h}{P_i}} = 0.0025$$

Simplify the equation:

$$\frac{P_i - P_f}{P_f} = 0.0025$$

Given that the change in momentum is $P_0$, we have:

$$\frac{P_i - (P_i + P_0)}{P_i + P_0} = 0.0025$$

Solve for $P_i$:

$$\frac{-P_0}{P_i + P_0} = 0.0025$$

$$-P_0 = 0.0025(P_i + P_0)$$

$$-P_0 = 0.0025P_i + 0.0025P_0$$

$$-P_0 - 0.0025P_0 = 0.0025P_i$$

$$-1.0025P_0 = 0.0025P_i$$

$$P_i = -\frac{1.0025}{0.0025}P_0$$

$$P_i = -401P_0$$

Therefore, the initial momentum of the proton was 401$P_0$.

Therefore, the correct answer is option 'c'.

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