Question

If de- Broglie wavelength of electron is equal to de- Broglie wavelength of proton, then what is the relation between their kinetic energy

• A. $KE_e > KE_p $
• B. $KE_p > KE_e $
• C. $KE_p = KE_e $
• D. $2KE_e = KE_p $

Answer 1

The Correct Option is A

To solve this question, we need to understand the concept of de Broglie wavelength and how it relates to the kinetic energy of particles.

The de Broglie wavelength is given by the formula:

\(\lambda = \frac{h}{p}\) 

where:

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

The momentum \(p\) can also be expressed as:

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

where:

• \(m\) is the mass of the particle.
• \(K\) is the kinetic energy of the particle.

Given that the de Broglie wavelength of the electron is equal to that of the proton, we have:

\(\lambda_e = \lambda_p\)

Substituting the expression for the de Broglie wavelength, we get:

\(\frac{h}{\sqrt{2m_e K_e}} = \frac{h}{\sqrt{2m_p K_p}}\)

Upon simplifying, this can be written as:

\(\sqrt{\frac{K_e}{m_e}} = \sqrt{\frac{K_p}{m_p}}\)

Squaring both sides gives:

\(\frac{K_e}{m_e} = \frac{K_p}{m_p}\)

Rearranging the terms, we have:

\(K_e = K_p \cdot \frac{m_e}{m_p}\)

Since the mass of the electron \(m_e\) is much less than the mass of the proton \(m_p\), the factor \(\frac{m_e}{m_p}\) is a very small number.

Therefore, for equal de Broglie wavelengths, the kinetic energy of the electron \(K_e\) must be greater than that of the proton \(K_p\) to compensate for its lower mass.

Thus, the correct answer is \(\large{KE_e > KE_p}\).