To solve the problem, we need to find the ratio of the frequencies of light that can produce photoelectrons with a maximum kinetic energy of 1 eV from three photosensitive materials with work functions of 1 eV, 2 eV, and 3 eV respectively.
### Step-by-Step Solution:
1. **Understanding the Photoelectric Effect**:
The maximum kinetic energy (K.E.) of the emitted photoelectrons can be described by the equation:
\[
K.E. = h\nu - \phi
\]
where \( K.E. \) is the maximum kinetic energy of the emitted electrons, \( h \) is Planck's constant, \( \nu \) is the frequency of the incident light, and \( \phi \) is the work function of the material.
2. **Setting Up the Equations**:
For each material, we can set up the equation based on the given work functions and the maximum kinetic energy of 1 eV.
- For Material A (Work function \( \phi_1 = 1 \, \text{eV} \)):
\[
1 \, \text{eV} = h\nu_1 - 1 \, \text{eV}
\]
Rearranging gives:
\[
h\nu_1 = 1 \, \text{eV} + 1 \, \text{eV} = 2 \, \text{eV}
\]
- For Material B (Work function \( \phi_2 = 2 \, \text{eV} \)):
\[
1 \, \text{eV} = h\nu_2 - 2 \, \text{eV}
\]
Rearranging gives:
\[
h\nu_2 = 1 \, \text{eV} + 2 \, \text{eV} = 3 \, \text{eV}
\]
- For Material C (Work function \( \phi_3 = 3 \, \text{eV} \)):
\[
1 \, \text{eV} = h\nu_3 - 3 \, \text{eV}
\]
Rearranging gives:
\[
h\nu_3 = 1 \, \text{eV} + 3 \, \text{eV} = 4 \, \text{eV}
\]
3. **Finding the Frequencies**:
We now have:
\[
h\nu_1 = 2 \, \text{eV}, \quad h\nu_2 = 3 \, \text{eV}, \quad h\nu_3 = 4 \, \text{eV}
\]
4. **Calculating the Ratio of Frequencies**:
To find the ratio of the frequencies, we can divide each equation by \( h \):
\[
\nu_1 = \frac{2 \, \text{eV}}{h}, \quad \nu_2 = \frac{3 \, \text{eV}}{h}, \quad \nu_3 = \frac{4 \, \text{eV}}{h}
\]
Therefore, the ratio of the frequencies is:
\[
\frac{\nu_1}{\nu_2} : \frac{\nu_2}{\nu_3} = 2 : 3 : 4
\]
5. **Final Answer**:
The ratio of the respective frequencies of light that produce photoelectrons of maximum kinetic energy of 1 eV from each of the materials is:
\[
2 : 3 : 4
\]
