\(K_p\) \(>\) \(K_e\)
\(K_p\) \(<\) \(K_e\)
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:
\(\frac{h}{\sqrt{2m_p K_p}} = \frac{h}{\sqrt{2m_e K_e}}\)
From this, we can cancel out Planck's constant (\(h\)) and rearrange to find:
\(\sqrt{\frac{K_p}{m_p}} = \sqrt{\frac{K_e}{m_e}}\)
Squaring both sides gives:
\(\frac{K_p}{m_p} = \frac{K_e}{m_e}\)
Solving for \(K_p\), we have:
\(K_p = K_e \cdot \frac{m_p}{m_e}\)
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:
\(K_p \lt K_e\)
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:
\[K_p<K_e.\]
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

The de Broglie wavelengths of a proton and an α particle are \( \lambda \) and \( 2\lambda \) respectively. The ratio of the velocities of proton and α particle will be:
The de-Broglie wavelength of an electron is the same as that of a photon. If the velocity of the electron is 25% of the velocity of light, then the ratio of the K.E. of the electron to the K.E. of the photon will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)