Question

The de - Broglie wavelength of a particle of kinetic energy 'K' is \(\lambda\), the wavelength of the particle if its kinetic energy is

• A. \(\frac{\lambda}{2}\)
• B. \(\lambda\)
• C. \(4\lambda\)
• D. \(2\lambda\)

Answer 1

The Correct Option is D

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

Where:

• \(\lambda\) = de Broglie wavelength 
• h = Planck's constant
• p = momentum of the particle

The momentum of the particle can be related to its kinetic energy (\( K \)) as:

\( p = \sqrt{2mK} \)

Where:

• m = mass of the particle
• K = kinetic energy of the particle

Now, consider the situation where the kinetic energy of the particle is: \( K' = \frac{\lambda}{4} \)

We can calculate the new wavelength (\( \lambda' \)) using the same equation:

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

Substitute the expression for momentum (\( p' \)) in terms of the new kinetic energy (\( K' \)):

\( \lambda' = \frac{h}{\sqrt{2mK'}} \)

Substitute \( K' = \frac{\lambda}{4} \) into the above equation:

\( \lambda' = \frac{h}{\sqrt{2m \left( \frac{\lambda}{4} \right)}} \)

Simplifying the expression:

\( \lambda' = \frac{h}{\sqrt{2m\frac{\lambda}{8}}} = \frac{h}{\left(\sqrt{\frac{\lambda}{4}} \times \sqrt{2m}\right)} = \frac{2h}{\sqrt{\lambda \times 2m}} \)

We can see that the new wavelength \( \lambda' \) is equal to:

\( \lambda' = 2\lambda \)

Therefore, the correct option is: (D) \( 2\lambda \)

Answer 2

The de Broglie wavelength \( \lambda \) of a particle is related to its momentum \( p \) by the equation: \[ \lambda = \frac{h}{p} \] where \( h \) is Planck's constant.

The kinetic energy \( K \) of a particle with mass \( m \) and velocity \( v \) is given by: \[ K = \frac{1}{2}mv^2 \] The momentum \( p \) is given by \( p = mv \). We can express the kinetic energy in terms of momentum: \[ K = \frac{(mv)^2}{2m} = \frac{p^2}{2m} \] Solving for momentum \( p \): \[ p = \sqrt{2mK} \]

Substituting this expression for momentum into the de Broglie wavelength equation: \[ \mathbf{\lambda = \frac{h}{\sqrt{2mK}}} \] This shows that the de Broglie wavelength is inversely proportional to the square root of the kinetic energy: \[ \lambda \propto \frac{1}{\sqrt{K}} \] 

Let the initial kinetic energy be \( K_1 = K \) and the corresponding wavelength be \( \lambda_1 = \lambda \). So, \( \lambda = \frac{h}{\sqrt{2mK}} \).

The question asks for the wavelength of the particle if its kinetic energy changes. Although the question text is incomplete ("...if its kinetic energy is"), the options provided are wavelengths (\( \lambda/2, \lambda, 4\lambda, 2\lambda \)). This implies a specific change in kinetic energy leads to one of these wavelengths. Let's assume the new kinetic energy is \( K_2 \). The new wavelength is \( \lambda_2 \). \[ \lambda_2 = \frac{h}{\sqrt{2mK_2}} \] We can find the ratio: \[ \frac{\lambda_2}{\lambda_1} = \frac{h/\sqrt{2mK_2}}{h/\sqrt{2mK_1}} = \sqrt{\frac{K_1}{K_2}} \] \[ \lambda_2 = \lambda_1 \sqrt{\frac{K_1}{K_2}} \]

Let's assume the question intended to ask for the wavelength if the kinetic energy becomes \( K/4 \). In this case, \( K_2 = K/4 \) and \( K_1 = K \). \[ \lambda_2 = \lambda \sqrt{\frac{K}{K/4}} \] \[ \lambda_2 = \lambda \sqrt{4} \] \[ \mathbf{\lambda_2 = 2\lambda} \]

This result matches one of the options.

If the kinetic energy is \( K/4 \), the new wavelength is \( 2\lambda \).