Questions: A hydrogen atom absorbs a photon, and an electron transitions from energy level 1 to 3. For scientific notation answers, please enter 1.23 e 21 to represent 1.23 x 10^21 What is the electron's initial energy? What is the electron's final energy? Ephoton = J h = s fphoton = s cphoton = m / s lambdaphoton = m

Solution Steps

• The energy of an electron in a hydrogen atom at a given energy level \( n \) is given by the formula: \[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
• For the initial energy level \( n = 1 \): \[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]

• For the final energy level \( n = 3 \): \[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} \approx -1.51 \, \text{eV} \]

• The energy of the photon absorbed is the difference between the final and initial energy levels: \[ E_{\text{photon}} = E_3 - E_1 = (-1.51 \, \text{eV}) - (-13.6 \, \text{eV}) = 12.09 \, \text{eV} \]
• Convert the energy from electron volts to joules using the conversion factor \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \): \[ E_{\text{photon}} = 12.09 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} \approx 1.94 \times 10^{-18} \, \text{J} \]

Final Answer