The de Broglie wavelength (\(\lambda\)) of a particle is calculated using the formula: \[ \lambda = \frac{h}{m v} \] In this equation, \( h \) represents Planck’s constant, \( m \) is the particle's mass, and \( v \) is its velocity. Given that all three particles (electron, proton, and deuteron) share the same speed, the wavelength is inversely proportional to their respective masses: \[ \lambda \propto \frac{1}{m} \] The masses of these particles are ordered as follows: \( m_e \) (electron) is the least massive, \( m_p \) (proton) is more massive than the electron, and \( m_d \) (deuteron) is the most massive. Consequently, their de Broglie wavelengths are ordered in the reverse manner: \[ \lambda_e > \lambda_p > \lambda_d \] Therefore, the correct answer is (2).
A beam of light of wavelength \(\lambda\) falls on a metal having work function \(\phi\) placed in a magnetic field \(B\). The most energetic electrons, perpendicular to the field, are bent in circular arcs of radius \(R\). If the experiment is performed for different values of \(\lambda\), then the \(B^2 \, \text{vs} \, \frac{1}{\lambda}\) graph will look like (keeping all other quantities constant).