lines converge at 9.12times 10^-8m in the emission of the hydrogen atom Calculate the first ionization energy of hydrogen if k) mol Worked example 3

To calculate the first ionization energy of hydrogen, we need to determine the energy required to remove the outermost electron from a hydrogen atom in its ground state.<br /><br />The first ionization energy is the energy required to remove the outermost electron from a neutral hydrogen atom in its ground state (n=1). This process can be represented by the following equation:<br /><br />H(g) → H+(g) + e energy of the emitted photon corresponds to the energy difference between the ground state and the excited state of the hydrogen atom. In this case, the lines converge at $9.12\times 10^{-8}m$, which represents the wavelength of the emitted photon.<br /><br />To calculate the energy of the emitted photon, we can use the formula:<br /><br />E = hc/λ<br /><br />where E is the energy, h is Planck's constant ($6.626\times 10^{-34} J\cdot s$), c is the speed of light ($3.00\times 10^8 m/s$), and λ is the wavelength.<br /><br />Substituting the given values, we have:<br /><br />E = (6.626\times 10^{-34} J\cdot s) \times (3.00\times 10^8 m/s) / (9.12\times 10^{-8} m)<br /><br />E = 2.18\times 10^{-18} J<br /><br />Since the energy of the emitted photon corresponds to the energy required to remove the outermost electron from the hydrogen atom, the first ionization energy of hydrogen is equal to the energy of the emitted photon.<br /><br />Therefore, the first ionization energy of hydrogen is $2.18\times 10^{-18} J$.