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

To solve this problem, we start with the de Broglie wavelength formula for an electron:

\(\lambda = \frac{h}{p}\)

where \(\lambda\) is the de Broglie wavelength, \(h\) is Planck's constant, and \(p\) is the momentum.

Given that the change in wavelength is \(1\%\), denote the initial momentum as \(p_i\) and the final momentum after change as \(p_f = p_i + P\).

The initial wavelength is:

\(\lambda_i = \frac{h}{p_i}\)

The final wavelength is:

\(\lambda_f = \frac{h}{p_f} = \frac{h}{p_i + P}\)

According to the problem, the change in wavelength is \(1\%\) of the initial wavelength:

\(\lambda_f = \lambda_i (1 + 0.01)\)

Substituting for \(\lambda_f\) and \(\lambda_i\), we have:

\(\frac{h}{p_i + P} = \frac{h}{p_i} \times 1.01\)

Simplifying, we get:

\(\frac{1}{p_i + P} = \frac{1.01}{p_i}\)

Cross-multiplying gives:

\(p_i = 1.01(p_i + P)\)

Expanding and simplifying the equation:

\(p_i = 1.01p_i + 1.01P\)

Rearranging the terms:

\(0.01p_i = 1.01P\)

Solving for \(p_i\):

\(p_i = 101P\)

Considering round-off effects and practical implementation, the closest option is:

100 P

Answer 2

• 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.