WBJEE 2024 Physics & Chemistry Paper

To determine the nature of the graph of $B^2$ vs. $rac{1}{\lambda}$, we start by analyzing the given problem using the photoelectric effect and the behavior of electrons in a magnetic field. Photoelectric Effect and Kinetic EnergyThe energy of the incident light is given by:$E=\frac{h c}{\lambda}$where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the incident light.The kinetic energy ( $K . E$.) of the emitted photoelectrons, given the work function $\phi$, is:$\text{K.E.}=\frac{h c}{\lambda}-\phi$ Electron in a Magnetic FieldWhen these electrons enter a perpendicular magnetic field $B$, they experience a force that causes them to move in a circular path. The centripetal force required for this motion is provided by the Lorentz force:$rac{m v^2}{R}=e v B$ where:$m$ is the mass of the electron, $v$ is the velocity of the electron,$R$ is the radius of the circular path,$e$ is the charge of the electron.From the above equation, we can solve for $v$ :$v=\frac{e B R}{m}$ Relationship between Kinetic Energy and VelocityThe kinetic energy of the electrons can also be expressed as:$\text{K.E.} = \frac{1}{2} m v^2$Substituting $v=\frac{e B R}{m}$ into the kinetic energy equation, we get:$\frac{1}{2} m \left( \frac{e B R}{m} \right)^2 = \frac{1}{2} \frac{e^2 B^2 R^2}{m}$$\text{K.E.} = \frac{e^2 B^2 R^2}{2 m}$ Combining Both Expressions for Kinetic EnergyWe already have:$\text{K.E.}=\frac{h c}{\lambda}-\phi$Equating both expressions for kinetic energy:$\frac{e^2 B^2 R^2}{2 m} = \frac{h c}{\lambda} - \phi$Rearranging to isolate $B^2$ :$B^2 = \frac{2 m}{e^2 R^2} \left( \frac{h c}{\lambda} - \phi \right)$$B^2=\frac{2 m h c}{e^2 R^2} \cdot \frac{1}{\lambda} - \frac{2 m \phi}{e^2 R^2}$ Form of the GraphThis equation is in the form of a straight line $y=m x+c$ where:$y=B^2$$x=\frac{1}{\lambda}$$m=\frac{2 m h c}{e^2 R^2}$$c=-\frac{2 m \phi}{e^2 R^2}$ Thus, the graph of $B^2$ vs. $rac{1}{\lambda}$ will be a straight line with a positive slope and a negative $y$-intercept.The correct graph will be a straight line with a positive slope and negative $y$-intercept:Option C