Question

If the momentum of an electron is changed by P, then the de Broglie wavelength associated with it changes by \(1\%\). The initial momentum of the electron will be:

• A. 200 P
• B. 100 P
• C. 300 P
• D. 150 P

Answer 1

The Correct Option is B

• Change in momentum = \( P \)
• Percentage change in de Broglie wavelength = 1%

De Broglie Wavelength Formula:

\[ \lambda = \frac{h}{p} \]

where:
\( h \) = Planck's constant
\( p \) = momentum of electron

Calculating the Initial Momentum:

Let initial momentum = \( p_0 \)

Initial wavelength: \[ \lambda_0 = \frac{h}{p_0} \]

New momentum: \[ p' = p_0 + P \]

New wavelength: \[ \lambda' = \frac{h}{p_0 + P} \]

Percentage change in wavelength:

\[ \frac{\Delta \lambda}{\lambda_0} \times 100 = 1\% \]

\[ \frac{\lambda' - \lambda_0}{\lambda_0} \times 100 = 1 \]

\[ \left(\frac{\frac{h}{p_0 + P}}{\frac{h}{p_0}} - 1\right) \times 100 = 1 \]

\[ \left(\frac{p_0}{p_0 + P} - 1\right) \times 100 = 1 \]

\[ \frac{-P}{p_0 + P} \times 100 = 1 \]

\[ \frac{P}{p_0 + P} = 0.01 \]

\[ P = 0.01p_0 + 0.01P \]

\[ 0.99P = 0.01p_0 \]

The initial momentum of the electron was 100P.

Answer 2

\(\text{ The de Broglie wavelength is given by } \lambda = \frac{h}{p}, \text{ where } p \text{ is the momentum.} \\ \text{ A 1\% change in wavelength corresponds to a 1\% change in momentum.} \\ \text{ If the change in momentum is } P, \text{ the initial momentum } p_0 \text{ is given by:}\)

\[\frac{P}{p_0} = 0.01 \implies p_0 = 100 P\]