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).
Step 1: Energy of the Emitted Electrons
The energy of the emitted electrons is related to the wavelength of the incident light by the photoelectric equation:
\[E = \frac{hc}{\lambda} - \phi,\]
where \(E\) is the kinetic energy of the emitted electrons, \(h\) is Planck’s constant, \(c\) is the speed of light, and \(\phi\) is the work function.
Step 2: Relation Between Kinetic Energy and Magnetic Field
The kinetic energy \(E\) of the electrons is also related to the magnetic field \(B\) by the radius \(R\) of the circular path they follow:
\[E = \frac{eB^2R^2}{2m},\]
where \(e\) is the charge of the electron and \(m\) is the mass of the electron.
Step 3: Combining the Relations
Combining the above relations and solving for \(B^2\), we obtain:
\[B^2 \propto \frac{1}{\lambda}.\]
The Correct Answer is: Option 3