To find the wavelength of the emitted photon when an electron jumps from the third excited state to the ground state in a hydrogen atom, we can follow these steps:
### Step 1: Identify the energy levels
The ground state energy of the hydrogen atom is given as \( E_1 = -13.6 \, \text{eV} \). The energy of the nth level is given by the formula:
\[
E_n = -\frac{13.6}{n^2} \, \text{eV}
\]
where \( n \) is the principal quantum number.
### Step 2: Calculate the energy of the third excited state
The third excited state corresponds to \( n = 4 \) (since the ground state is \( n = 1 \), the first excited state is \( n = 2 \), the second excited state is \( n = 3 \), and the third excited state is \( n = 4 \)).
\[
E_4 = -\frac{13.6}{4^2} = -\frac{13.6}{16} = -0.85 \, \text{eV}
\]
### Step 3: Calculate the energy difference
The energy difference \( \Delta E \) when the electron transitions from \( n = 4 \) to \( n = 1 \) is:
\[
\Delta E = E_1 - E_4 = (-13.6) - (-0.85) = -13.6 + 0.85 = -12.75 \, \text{eV}
\]
Since we are interested in the energy emitted, we take the absolute value:
\[
\Delta E = 12.75 \, \text{eV}
\]
### Step 4: Convert energy to wavelength
Using the relationship between energy and wavelength:
\[
E = \frac{hc}{\lambda}
\]
where \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)) and \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)). Rearranging gives:
\[
\lambda = \frac{hc}{E}
\]
### Step 5: Convert energy from eV to Joules
To use the above formula, we need to convert the energy from eV to Joules. The conversion factor is \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[
E = 12.75 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 2.04 \times 10^{-18} \, \text{J}
\]
### Step 6: Calculate the wavelength
Now substituting the values into the wavelength formula:
\[
\lambda = \frac{(6.63 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{2.04 \times 10^{-18} \, \text{J}}
\]
Calculating this gives:
\[
\lambda = \frac{1.989 \times 10^{-25}}{2.04 \times 10^{-18}} \approx 9.75 \times 10^{-8} \, \text{m}
\]
### Step 7: Convert to Angstroms
To convert meters to Angstroms (1 Angstrom = \( 10^{-10} \, \text{m} \)):
\[
\lambda \approx 9.75 \times 10^{-8} \, \text{m} = 975 \, \text{nm} = 9750 \, \text{Å}
\]
### Final Answer
The wavelength of the emitted photon is approximately \( 975 \, \text{nm} \) or \( 9750 \, \text{Å} \).
