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.
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.
