To solve the problem, we need to find the maximum kinetic energy of photoelectrons emitted when light of wavelength 155 nm is used on the same metallic surface, given that the maximum kinetic energy of photoelectrons emitted with a wavelength of 620 nm is 1 eV.
### Step-by-Step Solution:
1. **Understand the relationship between energy, work function, and kinetic energy:**
The energy of the incident photons can be expressed as:
\[
E = K.E + \phi
\]
where \(E\) is the energy of the photon, \(K.E\) is the kinetic energy of the emitted photoelectron, and \(\phi\) is the work function of the metal.
2. **Calculate the energy of the photon for the first wavelength (620 nm):**
The energy of a photon is given by the formula:
\[
E = \frac{hc}{\lambda}
\]
where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength. Given \(hc = 1240 \, \text{eV-nm}\) and \(\lambda_1 = 620 \, \text{nm}\):
\[
E_1 = \frac{1240 \, \text{eV-nm}}{620 \, \text{nm}} = 2 \, \text{eV}
\]
3. **Use the known maximum kinetic energy to find the work function:**
From the first case, we know that the maximum kinetic energy \(K.E_1 = 1 \, \text{eV}\):
\[
2 \, \text{eV} = K.E_1 + \phi
\]
Substituting \(K.E_1\):
\[
2 \, \text{eV} = 1 \, \text{eV} + \phi
\]
Thus, the work function \(\phi\) is:
\[
\phi = 2 \, \text{eV} - 1 \, \text{eV} = 1 \, \text{eV}
\]
4. **Calculate the energy of the photon for the second wavelength (155 nm):**
Now, for the second case with \(\lambda_2 = 155 \, \text{nm}\):
\[
E_2 = \frac{1240 \, \text{eV-nm}}{155 \, \text{nm}} \approx 8 \, \text{eV}
\]
5. **Calculate the maximum kinetic energy for the second wavelength:**
Using the relationship established earlier:
\[
E_2 = K.E_2 + \phi
\]
Substituting the values we found:
\[
8 \, \text{eV} = K.E_2 + 1 \, \text{eV}
\]
Therefore, the maximum kinetic energy \(K.E_2\) is:
\[
K.E_2 = 8 \, \text{eV} - 1 \, \text{eV} = 7 \, \text{eV}
\]
### Final Answer:
The maximum kinetic energy of the photoelectron emitted when a wavelength of 155 nm is used is **7 eV**.
