To solve the problem, we need to determine the minimum wavelength of photons emitted by mercury atoms in a Frank-Hertz experiment. Here's a step-by-step solution:
### Step 1: Determine the energy lost by the electron
The energy of the electron before passing through the mercury vapor is given as \(5.6 \, \text{eV}\), and the energy after passing through is \(0.7 \, \text{eV}\).
The energy lost by the electron can be calculated as:
\[
\text{Energy lost} = \text{Initial energy} - \text{Final energy} = 5.6 \, \text{eV} - 0.7 \, \text{eV} = 4.9 \, \text{eV}
\]
### Step 2: Relate energy to wavelength
The energy of a photon can be related to its wavelength using the formula:
\[
E = \frac{hc}{\lambda}
\]
where:
- \(E\) is the energy of the photon,
- \(h\) is Planck's constant (\(4.135667696 \times 10^{-15} \, \text{eV s}\)),
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \(\lambda\) is the wavelength in meters.
### Step 3: Rearrange the formula to find wavelength
Rearranging the formula to solve for wavelength gives:
\[
\lambda = \frac{hc}{E}
\]
### Step 4: Substitute the values
We can now substitute the values into the equation. First, we need to convert the energy from eV to joules if necessary, but since we are using \(h\) in eV·s, we can directly use the energy in eV.
Substituting \(E = 4.9 \, \text{eV}\):
\[
\lambda = \frac{(4.135667696 \times 10^{-15} \, \text{eV s})(3 \times 10^8 \, \text{m/s})}{4.9 \, \text{eV}}
\]
### Step 5: Calculate the wavelength
Calculating the above expression:
\[
\lambda = \frac{(4.135667696 \times 10^{-15} \times 3 \times 10^8)}{4.9}
\]
\[
\lambda \approx \frac{1.240 \times 10^{-6}}{4.9} \approx 2.53 \times 10^{-7} \, \text{m} = 253 \, \text{nm}
\]
### Step 6: Final answer
The minimum wavelength of photons emitted by mercury atoms is approximately \(253 \, \text{nm}\).
