For wave concerned with proton, de-Broglie wavelength change by \(0.25 \%\). If its momentum changes by \(\mathrm{P}_{\mathrm{O}}\) initial momentum $=\ldots \ldots \ldots$ (A) \(100 \mathrm{P}_{\mathrm{O}}\) (B) \(\left\\{\mathrm{P}_{\mathrm{O}} / 400\right\\}\) (C) \(401 \mathrm{P}_{\mathrm{O}}\) (D) \(\left\\{\mathrm{P}_{\mathrm{O}} / 100\right\\}\)

The de-Broglie wavelength formula for a particle is given by: \[\lambda = \frac{h}{p}\] where \(h\) is the Planck's constant and \(p\) is the momentum of the particle.

We know that the initial de-Broglie wavelength is \(\lambda\) and the initial momentum is \(p\). After the change, the final de-Broglie wavelength becomes \(\lambda + 0.0025\lambda\) (0.25% increase) and the final momentum becomes \(p + P_o\). Now, using the de-Broglie wavelength formula, we can write the equations for initial and final wavelengths: For initial wavelength, \[\lambda = \frac{h}{p}\] For final wavelength, \[\lambda + 0.0025\lambda = \frac{h}{p + P_o}\]

First, simplify the equation for the final wavelength as: \[1.0025\lambda = \frac{h}{p + P_o}\] Now we can substitute the initial wavelength equation into the final wavelength equation: \[1.0025\frac{h}{p} = \frac{h}{p+ P_o}\] Divide both sides by \(h\): \[1.0025\frac{1}{p} = \frac{1}{p + P_o}\] To solve for \(p\), first get a common denominator on the left side of the equation: \[\frac{1.0025p}{p(p + P_o)} = \frac{p + P_o}{p(p + P_o)}\] Now we can equate the numerators: \[1.0025p = p + P_o\] Rearranging the equation to solve for \(p\) gives: \[p(1-1.0025)=-P_o\] Finally, solve for \(p\): \[p = \frac{P_o}{1.0025-1} = \frac{P_o}{0.0025} = 400P_o\] The initial momentum \(p\) is \(400P_o\), so the correct answer is (B).