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).
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).
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The Correct Option is C
Solution and Explanation
Phase 1: Electron Emission Energy
The emitted electron energy links to incident light wavelength through the photoelectric equation:
\[E = \frac{hc}{\lambda} - \phi,\]
where \(E\) signifies kinetic energy, \(h\) is Planck’s constant, \(c\) denotes the speed of light, and \(\phi\) is the work function.
Phase 2: Kinetic Energy and Magnetic Field Relationship
Electron kinetic energy \(E\) also relates to the magnetic field \(B\) via the circular path radius \(R\):
\[E = \frac{eB^2R^2}{2m},\]
where \(e\) is the electron charge and \(m\) is the electron mass.
Phase 3: Relation Combination
Combining the above relations and solving for \(B^2\) yields:
\[B^2 \propto \frac{1}{\lambda}.\]