a. State the function of each of the following parts of a modern x-ray tube: i. heater; ii. high tension source; iii. cooling fins.
b. State one reason for each of the following design features of a modern x-ray tube: i. the glass envelope is highly evacuated; ii. the target is a metal of very high melting point; iii. the cooling fins are located outside the glass envelope.
c. In a nuclear fission reaction, a nuclide \(^{235}U_{92}\) is bombarded with a neutron to produce \(^{93}Kr_{36}\) and \(^{141}Ba_{56}\) with additional neutrons, the energy involved in the process is Q.
[mass of \(^{235}U_{92}\) = 235.044 u, mass of \(^{93}Kr_ {36}\) = 91.898 u, mass of \(^{141}Ba_ {56}\) = 140.914 u,
mass of neutron = 1.009u, 1u = \(1.66 \times 10^{-27}\) kg, c = 3.0 × \(10^8ms^1\)]
i. Write down the balanced nuclear reaction equation for the process.
ii. State with reason whether Q is absorbed or released in the process.
iii. Calculate the value of Q in joules.
ai. The heater, often referred to as the filament, is responsible for emitting electrons when heated.
ii. The high tension source, or high voltage generator, provides the high voltage necessary to accelerate the electrons emitted from the filament towards the anode (target) of the X-ray tube.
iii. Cooling fins are designed to dissipate the heat generated during the operation of the X-ray tube.
bi. The X-ray tube is highly evacuated so that the accelerated electrons can get to their target without losing much of their energy.
ii. The target is a metal with a very high melting point to withstand the high temperature generated when the electrons strike the target.
iii. The cooling fins are placed outside the glass envelope to ensure that the heat generated within the X-ray tube is efficiently transferred to the external environment.
ci. \(^{235}U_{92} + ^1_0n → ^{93}Kr_{36} + ^{141}Ba_{56} + 2^1_0n + Q\)
ii. Energy, Q, is released because the total mass on the left-hand side (reactant) is more than the total mass on the right-hand side (product). So, the energy is released in the form of a reduction in total mass. This missing mass is known as the 'mass defect' and it accounts for the energy released. Also, it's induced fission and nuclear fission is a process in which an unstable nucleus splits into two other lighter nuclei together with several neutrons and is accompanied by the release of energy.
iii. E = ∆\(mc^2\) where ∆m is the mass defect.
Total mass on the LHS:
235.044 u + 1.009 u = 236.053 u
Total mass on the RHS:
91.898 u + 140.914 u + 2(1.009 u) = 234.83 u
Mass defect, ∆m = 236.053 u - 234.83 u = 1.223 u
∆m = 1.223 × 1.66 × \(10^{-27}\) = 2.03 × \(10^{-27}\) kg
So,
E = 2.03 × \(10^{-27}\) × (3 × \(10^8)^2\)
= 1.827 × \(10^{-10}\) J
ai. The heater, often referred to as the filament, is responsible for emitting electrons when heated.
ii. The high tension source, or high voltage generator, provides the high voltage necessary to accelerate the electrons emitted from the filament towards the anode (target) of the X-ray tube.
iii. Cooling fins are designed to dissipate the heat generated during the operation of the X-ray tube.
bi. The X-ray tube is highly evacuated so that the accelerated electrons can get to their target without losing much of their energy.
ii. The target is a metal with a very high melting point to withstand the high temperature generated when the electrons strike the target.
iii. The cooling fins are placed outside the glass envelope to ensure that the heat generated within the X-ray tube is efficiently transferred to the external environment.
ci. \(^{235}U_{92} + ^1_0n → ^{93}Kr_{36} + ^{141}Ba_{56} + 2^1_0n + Q\)
ii. Energy, Q, is released because the total mass on the left-hand side (reactant) is more than the total mass on the right-hand side (product). So, the energy is released in the form of a reduction in total mass. This missing mass is known as the 'mass defect' and it accounts for the energy released. Also, it's induced fission and nuclear fission is a process in which an unstable nucleus splits into two other lighter nuclei together with several neutrons and is accompanied by the release of energy.
iii. E = ∆\(mc^2\) where ∆m is the mass defect.
Total mass on the LHS:
235.044 u + 1.009 u = 236.053 u
Total mass on the RHS:
91.898 u + 140.914 u + 2(1.009 u) = 234.83 u
Mass defect, ∆m = 236.053 u - 234.83 u = 1.223 u
∆m = 1.223 × 1.66 × \(10^{-27}\) = 2.03 × \(10^{-27}\) kg
So,
E = 2.03 × \(10^{-27}\) × (3 × \(10^8)^2\)
= 1.827 × \(10^{-10}\) J
