Fermi function. (a) under equilibrium conditions, (i) what is the probability of an electron state being occupied if it is located at the fermi level? at t = 0 k, what is the probability of an electron state being occupied if it located (ii) above the fermi level (iii) below the fermi level (b) plot fermi function for ef=0.5 ev and show the probability of an occupied state δe above ef is equal to the probability of an empty state δe below ef so f(ef +δe) = 1 - f(ef -δe). (c) if the probability of a state being empty at ev - 1kt is equal to the probability of a state being filled at ev - 4kt, where is the fermi level located? (d) if ef is positioned at ec, determine the probability of finding electrons in states ec+3kt (a numerical answer is required). repeat your calculation for the case when ef is positioned at ev and the energy level is located at ev-3kt

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    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s