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:
The problem is approached using the de Broglie wavelength formula for an electron:
\(\lambda = \frac{h}{p}\)
Here, \(\lambda\) represents the de Broglie wavelength, \(h\) is Planck's constant, and \(p\) denotes momentum.
A 1% change in wavelength is given. Let the initial momentum be \(p_i\) and the final momentum be \(p_f = p_i + P\).
The initial wavelength is calculated as:
\(\lambda_i = \frac{h}{p_i}\)
The final wavelength is:
\(\lambda_f = \frac{h}{p_f} = \frac{h}{p_i + P}\)
The problem states that the final wavelength is 1% greater than the initial wavelength:
\(\lambda_f = \lambda_i (1 + 0.01)\)
Substituting the expressions for \(\lambda_f\) and \(\lambda_i\):
\(\frac{h}{p_i + P} = \frac{h}{p_i} \times 1.01\)
Simplifying this equation yields:
\(\frac{1}{p_i + P} = \frac{1.01}{p_i}\)
Cross-multiplication leads to:
\(p_i = 1.01(p_i + P)\)
Expanding and rearranging the terms:
\(p_i = 1.01p_i + 1.01P\)
Further rearrangement gives:
\(0.01p_i = 1.01P\)
Solving for \(p_i\):
\(p_i = 101P\)
Considering potential rounding and practical applications, the closest approximation is:
100 P