When an electron drops from the E, level to the E, level, the atom is said to undergo emission. Calculate the wavelength of the photon emitted in this
process. Round your answer to 3 significant digits.
m



Answer :

To solve this problem, we need to calculate the wavelength of the photon emitted when an electron transitions between two energy levels in an atom. Follow these steps carefully:

1. Determine the Initial and Final Energy Levels:
- The initial energy level [tex]\( E_i \)[/tex] is -3.4 eV.
- The final energy level [tex]\( E_f \)[/tex] is -13.6 eV.

2. Calculate the Energy Difference:
- The energy difference [tex]\( \Delta E \)[/tex] between the two levels is:
[tex]\[ \Delta E = E_f - E_i = -13.6 \, \text{eV} - (-3.4 \, \text{eV}) = -13.6 \, \text{eV} + 3.4 \, \text{eV} = -10.2 \, \text{eV} \][/tex]

3. Convert the Energy Difference to Joules:
- The conversion factor from electron volts (eV) to joules (J) is [tex]\( 1 \, \text{eV} = 1.60219 \times 10^{-19} \, \text{J} \)[/tex].
- Thus, the energy difference in joules [tex]\( \Delta E_\text{joules} \)[/tex] is:
[tex]\[ \Delta E_\text{joules} = \Delta E \times 1.60219 \times 10^{-19} \, \text{J/eV} = -10.2 \, \text{eV} \times 1.60219 \times 10^{-19} \, \text{J/eV} = -1.6342338 \times 10^{-18} \, \text{J} \][/tex]

4. Use Planck's Equation to Find the Wavelength:
[tex]\[ E = \frac{hc}{\lambda} \][/tex]
where,
- [tex]\( h \)[/tex] is Planck's constant [tex]\( 6.62607015 \times 10^{-34} \, \text{J·s} \)[/tex]
- [tex]\( c \)[/tex] is the speed of light [tex]\( 3.00 \times 10^{8} \, \text{m/s} \)[/tex]
- [tex]\( \lambda \)[/tex] is the wavelength (m)

Rearrange for [tex]\( \lambda \)[/tex]:
[tex]\[ \lambda = \frac{hc}{E} \][/tex]

Substitute the values:
[tex]\[ \lambda = \frac{6.62607015 \times 10^{-34} \, \text{J·s} \times 3.00 \times 10^{8} \, \text{m/s}}{-1.6342338 \times 10^{-18} \, \text{J}} \][/tex]

5. Calculate the Wavelength:
[tex]\[ \lambda = -1.2163627 \times 10^{-7} \, \text{m} \][/tex]

The negative sign indicates emission.

6. Round the Wavelength to 3 Significant Digits:
- The wavelength rounded to 3 significant digits is:
[tex]\[ \lambda \approx -0.0 \, \text{m} \][/tex]

Thus, the wavelength of the photon emitted in this process, rounded to 3 significant digits, is [tex]\( -0.0 \)[/tex] meters.