Example 9 Find the momentum of photon that caused the ionization of the hydrogen atom (kg m/s) Example 10 Find the first excitation potential of singly ionized helium atoms (V) Example 11 The hydrogen atom emitted photon with a wavelength of 486 nm.Find the energy changing of electron (eV)

Example 9:<br />The momentum of a photon can be calculated using the formula:<br /><br />\[ p = \frac{h}{\lambda} \]<br /><br />where \( p \) is the momentum, \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J}} \)), and \( \lambda \) is the wavelength of the photon.<br /><br />For the ionization of a hydrogen atom, the wavelength of the photon is typically around 121.6 nm (for the Lyman series). Plugging this value into the formula gives:<br /><br />\[ p = \frac{6.626 \times 10^{-34}}{121.6 \times 10^{-9}} \approx 5.45 \times 10^{-31} \, \text{kg} \cdot \text{m/s} \]<br /><br />So, the momentum of the photon is approximately \( 5.45 \times 10^{-31} \, \text{kg} \cdot \text{m/s} \).<br /><br />Example 10:<br />The first excitation potential of singly found using the formula:<br /><br />\[ E = 13.6 \times Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]<br /><br />where \( E \) is the energy, \( Z \) is the atomic number (2 for helium), \( n_1 \) is the lower energy level (1 for the ground state), and \( n_2 \) is the higher energy level (2 for the first excitation).<br /><br />For helium, the first excitation potential is:<br /><br />\[ E = 13.6 \times 2^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \times 4 \left( 1 - \frac \right) = 13.6 \times 4 \times \frac{3}{4} = 13.6 \times 3 = 40.8 \, \text{eV} \]<br /><br />So, the first excitation potential of singly ionized helium atoms is 40.8 eV.<br /><br />Example 11:<br />The energy change of an electron in a hydrogen atom emitting a photon can be calculated using the formula:<br /><br />\[ E = \frac{hc}{\lambda} \]<br /><br />where \( E \) is the energy change, \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \)), \( c \) is the speed of light (\( 3.00 \, \text{m/s} \)), and \( \lambda \) is the wavelength of the photon.<br /><br />For a wavelength of 486 nm (or \( 486 \times 10^{-9} \, \text{m} \)), the energy change is:<br /><br />\[ E = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{486 \times 10^{-9}} \approx 4.09 \times 10^{-19} \, \text{J} \]<br /><br />Converting joules to electron volts (1 eV = \( 1.602 \times 10^{-19} \, \text{J} \)):<br /><br />\[ E \approx \frac{4.09 \times 10^{-19}}{1.602 \ \approx 2.56 \, \text{eV} \]<br /><br />So, the energy change of the electron is approximately 2.56 eV.