Question

The de-Broglie wavelength of a particle having kinetic energy E is \(\lambda\). How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value ?

• A. E
• B. \( \frac{1}{9} \) E
• C. \( \frac{7}{9} \) E
• D. \( \frac{16}{9} \) E

Hint:

For problems involving ratios and percentage changes, focusing on the proportionality between quantities is very efficient. Here, knowing \( \lambda \propto E^{-1/2} \) allows you to set up the ratio \( \lambda_2 / \lambda_1 = (E_2 / E_1)^{-1/2} = \sqrt{E_1 / E_2} \) directly, saving time on rewriting the full formulas. Pay close attention to what the question asks for - in this case, "extra energy" (\(E_2-E_1\)), not the final energy (\(E_2\)).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given the relationship between the de-Broglie wavelength and kinetic energy. We need to find the additional kinetic energy required to reduce the wavelength to a specific fraction of its original value.
Step 2: Key Formula or Approach:
The de-Broglie wavelength (\(\lambda\)) is related to momentum (p) by \( \lambda = h/p \).
The kinetic energy (E) is related to momentum by \( E = p^2 / (2m) \), which means \( p = \sqrt{2mE} \).
Combining these, we get the relationship between wavelength and kinetic energy:
\[ \lambda = \frac{h}{\sqrt{2mE}} \] This shows that \( \lambda \propto \frac{1}{\sqrt{E}} \).
Step 3: Detailed Explanation:
Let the initial state be denoted by subscript 1 and the final state by subscript 2.
Initial state: Wavelength = \(\lambda_1 = \lambda\), Kinetic Energy = \(E_1 = E\).
Final state: Wavelength = \(\lambda_2\), Kinetic Energy = \(E_2\).
We are given that the wavelength reduces to 75% of the initial value:
\[ \lambda_2 = 0.75 \lambda_1 = \frac{3}{4} \lambda \] From the relationship \( \lambda \propto 1/\sqrt{E} \), we can write:
\[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{E_1}{E_2}} \] Substitute the given values:
\[ \frac{(3/4)\lambda}{\lambda} = \sqrt{\frac{E}{E_2}} \] \[ \frac{3}{4} = \sqrt{\frac{E}{E_2}} \] Square both sides to solve for \(E_2\):
\[ \left(\frac{3}{4}\right)^2 = \frac{E}{E_2} \implies \frac{9}{16} = \frac{E}{E_2} \] \[ E_2 = \frac{16}{9} E \] The question asks for the "extra energy" (\(\Delta E\)) that must be given. This is the difference between the final and initial energies.
\[ \Delta E = E_2 - E_1 = \frac{16}{9} E - E \] \[ \Delta E = \left(\frac{16}{9} - 1\right) E = \left(\frac{16 - 9}{9}\right) E = \frac{7}{9} E \] Step 4: Final Answer:
The extra energy required is \( \frac{7}{9} E \). This corresponds to option (C).