Example 6 A photon corresponding to the first line of the Lyman series of the He ion knocks out an electron from a hydrogen atom at rest Find the kinetic energy of the ejected electron (eV)

To solve this problem, we need to use the conservation of energy principle. The energy of the incoming photon is equal to the sum of the kinetic energy of the ejected electron and the binding energy of the hydrogen atom.<br /><br />The Lyman series corresponds to transitions to the n=1 energy level in hydrogen. The first line of the Lyman series corresponds to the transition from n=2 to n=1. The energy of the photon corresponding to this transition can be calculated using the Rydberg formula:<br /><br />E = 13.6 eV * (1/n1^2 - 1/n2^2)<br /><br />where n1 and n2 are the principal quantum numbers of the initial and final energy levels, respectively. For the first line of the Lyman series, n1 = 1 and n2 = 2, so:<br /><br />E = 13.6 eV * (1/1^2 - 1/2^2) = 10.2 eV<br /><br />The binding energy of the hydrogen atom is 13.6 eV. Therefore, the kinetic energy of the ejected electron is:<br /><br />KE = E - 13.6 eV = 10.2 eV - 13.6 eV = -3.4 eV<br /><br />The negative sign indicates that the kinetic energy of the ejected electron is less than the energy of the incoming photon. Therefore, the kinetic energy of the ejected electron is 3.4 eV.