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

To determine how much extra energy is required to reduce the de-Broglie wavelength to 75% of its initial value, we need to understand the relationship between kinetic energy and de-Broglie wavelength.

The de-Broglie wavelength (\lambda) of a particle is given by the equation:

\lambda = \frac{h}{\sqrt{2mE}}

where h is Planck's constant, m is the mass of the particle, and E is its kinetic energy.

Initially, when the kinetic energy is E, the wavelength is \lambda. If the wavelength is reduced to 75% of the initial value, the new wavelength is \lambda' = 0.75 \lambda.

According to the de-Broglie equation,

\lambda' = \frac{h}{\sqrt{2mE'}}

where E' is the new kinetic energy.

We equate for the two scenarios:

\frac{h}{\sqrt{2mE'}} = 0.75 \times \frac{h}{\sqrt{2mE}}

Cancel h from both sides, and rearrange to get:

\sqrt{2mE'} = \frac{4}{3} \times \sqrt{2mE}

Square both sides to eliminate the square root:

2mE' = \left(\frac{4}{3}\right)^2 \times 2mE

E' = \frac{16}{9}E

To find the extra energy required, calculate the difference between E' and E:

\Delta E = E' - E = \frac{16}{9}E - E = \frac{16}{9}E - \frac{9}{9}E = \frac{7}{9}E

This matches the correct answer: the extra energy required is \frac{7}{9}E.