Question

If de-Broglie wavelength of electron is increased by \(1 \%\) its momentum \(\ldots \ldots\) (A) increases by \(1 \%\) (B) decreases by \(1 \%\) (C) increased by \(2 \%\) (D) decreases by \(2 \%\)

Short answer

The correct answer is (B) decreases by 1%. The momentum of the electron decreases due to the increase in its de-Broglie wavelength by 1%. Using the de-Broglie wavelength formula \(\lambda = \frac{h}{p}\), we find that \(p' = \frac{100}{101} \cdot p\), resulting in a percentage change in momentum of \(-\frac{100}{101}\%\).

Step-by-step solution

Write down the de-Broglie wavelength formula

The de-Broglie wavelength formula is as follows: \(\lambda = \frac{h}{p}\) Here, \(\lambda\) represents the de-Broglie wavelength, \(h\) is the Planck's constant, and \(p\) is the momentum of the particle.

Calculate the new wavelength

Since the wavelength (λ) is increased by 1%, the new wavelength becomes: \(\lambda' = \lambda (1 + \frac{1}{100}) = \lambda \cdot \frac{101}{100}\)

Find the new momentum

We have the formula \(\lambda = \frac{h}{p}\), so for the new wavelength and momentum, we get: \(\lambda' = \frac{h}{p'}\) Now, insert the value of \(\lambda'\) into this equation: \(\lambda \cdot \frac{101}{100} =\frac{h}{p'}\)

Express p' in terms of p

We can rearrange the equation in Step 3 to find p': \(p' = \frac{100}{101} \cdot \frac{h}{\lambda}\) Now, use the original de-Broglie formula \(\lambda = \frac{h}{p}\) to substitute h/λ with p: \(p' = \frac{100}{101} \cdot p\)

Calculate the percentage change in momentum

We can now calculate the percentage change in p: \(\frac{p' - p}{p} \times 100 = \frac{\frac{100}{101}p - p}{p} \times 100 = -\frac{1}{101} \times 100 = -\frac{100}{101}\%\) Since the percentage change in momentum is negative, the momentum decreases. Given the options, we can now conclude that the correct answer is: (B) decreases by 1%.