A proton and an electron have the same de Broglie wavelength. If \(K_p\) and \(K_e\) be the kinetic energies of proton and electron respectively. Then choose the correct relation:
To solve this problem, we need to understand the concept of de Broglie wavelength and how it relates to the kinetic energy of particles like protons and electrons.
The de Broglie wavelength (\(\lambda\)) of a particle is given by the equation:
\(\lambda = \frac{h}{p}\)
where \(h\) is Planck's constant and \(p\) is the momentum of the particle.
The momentum (\(p\)) can also be related to kinetic energy (\(K\)) using the relation:
\(p = \sqrt{2mK}\)
where \(m\) is the mass of the particle.
Given that the de Broglie wavelengths of a proton and an electron are the same, we can set up the equality:
Since the mass of the proton (\(m_p\)) is much greater than the mass of the electron (\(m_e\)), it follows that:
\(K_p \lt K_e\)
This shows that the kinetic energy of the proton is less than the kinetic energy of the electron when both have the same de Broglie wavelength.
Thus, the correct answer is:
\(K_p \lt K_e\)
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Approach Solution -2
The de Broglie wavelength for a particle is given by: \[\lambda = \frac{h}{p},\] where \(h\) is Planck's constant and \(p\) is the momentum. For the proton and electron: \[\lambda_{\text{proton}} = \lambda_{\text{electron}} \implies p_{\text{proton}} = p_{\text{electron}}.\] The kinetic energy is related to momentum as: \[K = \frac{p^2}{2m}.\] Since \(p_{\text{proton}} = p_{\text{electron}}\): \[K_{\text{proton}} = \frac{p^2}{2m_p}, \quad K_{\text{electron}} = \frac{p^2}{2m_e}.\] Given \(m_p>m_e\), it follows that: \[K_{\text{proton}}<K_{\text{electron}}.\] Thus: \[K_p<K_e.\]
