Question

If the linear momentum of a proton is changed by \(p_0\), then the de Broglie wavelength associated with the proton changes by 0.25%. Then the initial linear momentum of the proton is

• A. \(100 p_0\)
• B. \(\frac{p_0}{400}\)
• C. \(400 p_0\)
• D. \(\frac{p_0}{100}\)

Hint:

Small relative changes in momentum inversely affect de Broglie wavelength.

Answer 1

The Correct Option is C

Step 1: de Broglie wavelength relation
\[ \lambda = \frac{h}{p} \] where \(p\) is momentum.
Step 2: Calculate change in wavelength
Given: \[ \frac{\Delta \lambda}{\lambda} = 0.25% = 0.0025 \] Since \(\lambda = \frac{h}{p}\), small change in \(\lambda\) due to change in \(p\) is: \[ \frac{\Delta \lambda}{\lambda} = \frac{\Delta p}{p} \] But as momentum changes by \(p_0\), \(\Delta p = p_0\). So: \[ 0.0025 = \frac{p_0}{p} \implies p = \frac{p_0}{0.0025} = 400 p_0 \] Step 3: Conclusion
Initial momentum is \(400 p_0\).