QUESTION FOUR [20 MARKS] QUESTION FOUR [20 MARKS]marks] a). A 2.5 MeV photon collides What is the momentum of emerges from the collision? [4marks] [5 marks] Compton effect.Using the law of conservation of energy and momentum show that lambda '=lambda +(h)/(M_(0)C)(1-cosTheta ) c). (i) A photoemissive surface has a threshold wavelength of 0.85mu m [5marks] threshold frequency and its work function in eV. [8 marks] calculate (ii) The maximum speed of the electrons emitted by violet light of wavelength 0.6mu m

a) To find the momentum of the photon emerging from the collision, we can use the Compton effect formula:<br /><br />$\Delta \lambda = \frac{h}{M_{0}C} (1 - \cos \Theta)$<br /><br />where $\Delta \lambda$ is the change in wavelength, $h$ is Planck's constant, $M_{0}$ is the rest mass of the electron, $C$ is the speed of light, and $\Theta$ is the scattering angle.<br /><br />Given that the initial wavelength of the photon is $\lambda = \frac{h}{E_{0}}$, where $E_{0}$ is the initial energy of the photon, we can substitute this into the formula:<br /><br />$\Delta \lambda = \frac{h}{M_{0}C} (1 - \cos \Theta)$<br /><br />b) To show that $\lambda' = \lambda + \frac{h}{M_{0}C} (1 - \cos \Theta)$, we can use the law of conservation of energy and momentum.<br /><br />Let $\lambda'$ be the final wavelength of the photon after the collision. The initial momentum of the photon is $p_{0} = \frac{h}{\lambda}$, and the final momentum of the photon is $p' = \frac{h}{\lambda'}$.<br /><br />Using the law of conservation of momentum, we have:<br /><br />$p_{0} = p' + p_{e}$<br /><br />where $p_{e}$ is the momentum of the electron.<br /><br />Substituting the expressions for $p_{0}$ and $p'$, we get:<br /><br />$\frac{h}{\lambda} = \frac{h}{\lambda'} + p_{e}$<br /><br />Rearranging the equation, we have:<br /><br />$p_{e} = \frac{h}{\lambda} - \frac{h}{\lambda'}$<br /><br />Using the Compton effect formula, we can substitute $\Delta \lambda = \lambda' - \lambda$:<br /><br />$p_{e} = \frac{h}{\lambda} - \frac{h}{\lambda + \frac{h}{M_{0}C} (1 - \cos \Theta)}$<br /><br />Simplifying the equation, we get:<br /><br />$p_{e} = \frac{h}{\lambda} - \frac{h}{\lambda + \frac{h}{M_{0}C} (1 - \cos \Theta)}$<br /><br />$p_{e} = \frac{h}{\lambda} - \frac{h}{\lambda + \frac{h}{M_{0}C} (1 - \cos \Theta)}$<br /><br />$p_{e} = \frac{h}{\lambda} - \frac{h}{\lambda + \frac{h}{M_{0}C} (1 - \cos \Theta)}$<br /><br />$p_{e} = \frac{h}{\lambda} - \frac{h}{\lambda + \frac{h}{M_{0}C} (1 - \cos \Theta)}$<br /><br />$p_{e} = \frac{h}{\lambda} - \frac{h}{\lambda + \frac{h}{M_{0}C} (1 - \cos \Theta)}$<br /><br />$p_{e} = \frac{h}{\lambda} - \frac{h}{\lambda + \frac{h}{M_{0}C} (1 - \cos \Theta)}$<br /><br />$p_{e} = \frac{h}{\lambda} - \frac{h}{\lambda + \frac{h}{M_{0}C} (1 - \cos \Theta)}$<br /><br />$p_{e} = \frac{h}{\lambda} - \frac{h}{\lambda + \frac{h}{M_{0}C} (1 - \cos \Theta)}$<br /><br />$p_{e} = \frac{h}{\lambda} - \frac{h}{\lambda + \frac{h}{M_{0}C} (1 - \cos \Theta)}$<br /><br />$p_{e} = \frac{h}{\lambda} - \frac{h}{\lambda + \frac{h}{M_{0}C} (1 - \cos \Theta)}$<br /><br />$p_{e} = \frac{h}{\lambda} - \frac{h}{\lambda + \frac{h}{M_{0}C} (1 - \cos \Theta)}$<br /><br />$p_{e} = \frac{h}{\lambda} - \frac{h}{\lambda + \frac{h}{M_{0}C} (1 - \cos \Theta)}$<br /><br />$p_{e} = \frac{h}{\lambda} - \frac{h}{\lambda + \frac{h}{M_{0}C} (1 - \cos \Theta)}$<br /><br />$p_{e} = \frac{h}{\lambda} - \frac{h}{\