(f) Estimate the energy of the characteristic X-ray emitted from a tungsten target when an electron drops from a N shell (n=4) to a vacancy in the K shell (n=1)cdot Z=74 . (5 marks) (g) Draw an energy level diagram for hydrogen and at least show four series. (4 marks) (h) The nuclear reaction (}_{0)^1n+_(5)^10Barrow _(3)^7Li+_(2)^4He is observed to occur even when very slow moving neutrons (Mn=1.0087u) strike a boron atom at rest. For a particular reaction in which KE=0 , the helium (M_(He)=4.0026u) is observed to have a speed of 9.30ast 106m/s Determine (i) The KE of the lithium (M_(Li)=7.0160) (ii) The Q value of the reaction Page 1 of 3 PHYS 414 (4 marks) (2 marks)

(f) To estimate the energy of the characteristic X-ray emitted from a tungsten target when an electron drops from a N shell $(n=4)$ to a vacancy in the K shell $(n=1)$, we can use the formula:<br /><br />$E = 13.6 \text{ keV} \times Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$<br /><br />where $Z$ is the atomic number of tungsten (74), $n_f$ is the final energy level (1), and $n_i$ is the initial energy level (4).<br /><br />Plugging in the values, we get:<br /><br />$E = 13.6 \text{ keV} \times 74^2 \left( \frac{1}{1^2} - \frac{1}{4^2} \right)$<br /><br />Simplifying the expression, we find:<br /><br />$E = 13.6 \text{ keV} \times 5476 \left( 1 - \frac{1}{16} \right)$<br /><br />$E = 13.6 \text{ keV} \times 5476 \times \frac{15}{16}$<br /><br />$E = 13.6 \text{ keV} \times 5476 \times 0.9375$<br /><br />$E = 74.8 \text{ keV}$<br /><br />Therefore, the energy of the characteristic X-ray emitted from the tungsten target when an electron drops from the N shell $(n=4)$ to a vacancy in the K shell $(n=1)$ is approximately 74.8 keV.<br /><br />(g) To draw an energy level diagram for hydrogen, we need to show at least four series. The energy levels for hydrogen are given by the formula:<br /><br />$E_n = -13.6 \text{ eV} \times \frac{1}{n^2}$<br /><br />where $n$ is the principal quantum number.<br /><br />The four series that can be shown in the energy level diagram for hydrogen are:<br /><br />1. Lyman series: Transitions from higher energy levels to the n=1 level.<br />2. Balmer series: Transitions from higher energy levels to the n=2 level.<br />3. Paschen series: Transitions from higher energy levels to the n=3 level.<br />4. Brackett series: Transitions from higher energy levels to the n=4 level.<br /><br />The energy level diagram for hydrogen should show these four series with their respective wavelengths and energy levels.<br /><br />(h) To determine the kinetic energy (KE) of the lithium nucleus and the Q value of the reaction, we can use the principles of conservation of momentum and energy.<br /><br />(i) The kinetic energy of the lithium nucleus can be calculated using the formula:<br /><br />$KE = \frac{1}{2} m v^2$<br /><br />where $m$ is the mass of the lithium nucleus and $v$ is its velocity.<br /><br />Given that the mass of the lithium nucleus is 7.0160 u and the velocity is $9.30 \times 10^6$ m/s, we can substitute these values into the formula to find the kinetic energy:<br /><br />$KE = \frac{1}{2} \times 7.0160 \text{ u} \times (9.30 \times 10^6 \text{ m/s})^2$<br /><br />Converting the mass to kilograms and calculating the kinetic energy, we get:<br /><br />$KE = 0.007016 \text{ kg} \times (9.30 \times 10^6 \text{ m/s})^2$<br /><br />$KE = 6.72 \times 10^6 \text{ J}$<br /><br />(ii) The Q value of the reaction can be calculated using the formula:<br /><br />$Q = \text{Total kinetic energy of the products} - \text{Total kinetic energy of the reactants}$<br /><br />In this case, the total kinetic energy of the products is the sum of the kinetic energy of the lithium nucleus and the kinetic energy of the helium nucleus. The total kinetic energy of the reactants is the kinetic energy of the neutron.<br /><br />Given that the kinetic energy of the helium nucleus is 0.5 MeV (converted to Joules), we can calculate the Q value:<br /><br />$Q = (6.72 \times 10^6 \text{ J} + 0.5 \text{ MeV} \times 1.602 \times 10^{-13} \text{ J}) - 1.0087 \text{ u} \times 931.5 \text{ MeV/u}$<br /><br />Simplifying the expression, we find:<br /><br />$Q = (6.72 \times 10^6 \text{ J} + 8.1 \times 10^{-13} \text{ J}) - 1.0087 \times 931.5 \times 1.6605 \times