If the work functions of three photosensitive materials are 1 eV, 2 eV and 3 eV respectively, then the ratio of the respective frequencies of light that produce photoelectrons of the maximum kinetic energy of 1 eV from each of them is

Options

A. `1:2:3`
B. `2:3:4`
C. `1:1:1`
D. `3:2:1`

Answer

To solve the problem, we need to determine the ratio of the frequencies of light that can produce photoelectrons with a maximum kinetic energy of 1 eV from three different photosensitive materials with work functions of 1 eV, 2 eV, and 3 eV, respectively.

### Step-by-step Solution:

1. **Understand the relationship between kinetic energy, frequency, and work function**:
The maximum kinetic energy (KE_max) of the emitted photoelectrons can be expressed using the equation:
\[ KE_{max} = h\nu - \phi \]
where \( h \) is Planck's constant, \( \nu \) is the frequency of the incident light, and \( \phi \) is the work function of the material.

2. **Set up the equation for each material**:
For each material, we will set the maximum kinetic energy to 1 eV and solve for the frequency \( \nu \).

- **Material 1** (Work function \( \phi_1 = 1 \, \text{eV} \)):
\[ 1 = h\nu_1 - 1 \]
Rearranging gives:
\[ h\nu_1 = 1 + 1 = 2 \, \text{eV} \quad \Rightarrow \quad \nu_1 = \frac{2}{h} \]

- **Material 2** (Work function \( \phi_2 = 2 \, \text{eV} \)):
\[ 1 = h\nu_2 - 2 \]
Rearranging gives:
\[ h\nu_2 = 1 + 2 = 3 \, \text{eV} \quad \Rightarrow \quad \nu_2 = \frac{3}{h} \]

- **Material 3** (Work function \( \phi_3 = 3 \, \text{eV} \)):
\[ 1 = h\nu_3 - 3 \]
Rearranging gives:
\[ h\nu_3 = 1 + 3 = 4 \, \text{eV} \quad \Rightarrow \quad \nu_3 = \frac{4}{h} \]

3. **Find the ratio of the frequencies**:
Now we can find the ratio of the frequencies \( \nu_1 : \nu_2 : \nu_3 \):
\[ \nu_1 : \nu_2 : \nu_3 = \frac{2}{h} : \frac{3}{h} : \frac{4}{h} \]
Since \( h \) is a common factor, it cancels out:
\[ \nu_1 : \nu_2 : \nu_3 = 2 : 3 : 4 \]

### 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:
\[ \boxed{2 : 3 : 4} \]