Question

The energy levels of an atom are shown in the figure. Which one of these transitions will result in the emission of a photon of wavelength 124.1 nm? Given \( h = 6.62 \times 10^{-34} \) Js. 

• A. B
• B. A
• C. C
• D. D

Hint:

The energy of the photon is calculated using \( E = \frac{h c}{\lambda} \), and the correct transition is determined by the energy difference between the levels.

Answer 1

The Correct Option is D

The energy of a photon emitted during a transition is related to the wavelength of the photon by the equation: \[ E = \frac{h c}{\lambda} \] where: - \( E \) is the energy of the photon, - \( h \) is Planck’s constant (\( 6.62 \times 10^{-34} \, {Js} \)), - \( c \) is the speed of light (\( 3 \times 10^8 \, {m/s} \)), - \( \lambda \) is the wavelength of the photon. 
Step 1: Calculate the energy of the photon. Given that \( \lambda = 124.1 \, {nm} = 124.1 \times 10^{-9} \, {m} \), we can substitute the values into the equation: \[ E = \frac{(6.62 \times 10^{-34}) (3 \times 10^8)}{124.1 \times 10^{-9}} \] \[ E = \frac{1.986 \times 10^{-25}}{124.1 \times 10^{-9}} = 1.6 \times 10^{-18} \, {J} \] 
Step 2: Convert energy from joules to electron volts. Since \( 1 \, {eV} = 1.6 \times 10^{-19} \, {J} \), we convert the energy: \[ E = \frac{1.6 \times 10^{-18}}{1.6 \times 10^{-19}} = 10 \, {eV} \] 
Step 3: Check the energy differences between the levels. Now, we check the energy differences between the levels: 
- \( {A to B}: 0 - (-2.2) = 2.2 \, {eV} \) - \( {A to C}: 0 - (-5.2) = 5.2 \, {eV} \) - \( {A to D}: 0 - (-10) = 10 \, {eV} \) The transition from level A to level D gives the energy of 10 eV, which matches the energy of the photon calculated. 
Thus, the correct transition is \( A \rightarrow D \).