What is the threshold wavelength for a cesium surface, for which the work function is 1.8 eV ?

Explanation

To find the threshold wavelength for a cesium surface with a work function of 1.8 eV, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Work Function Equation**:
The work function (φ) is related to the threshold wavelength (λ₀) by the equation:
\[ \phi = \frac{hc}{\lambda_0} \]
where:
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda_0 \) is the threshold wavelength.

2. **Rearrange the Equation**:
To find the threshold wavelength, we can rearrange the equation:
\[ \lambda_0 = \frac{hc}{\phi} \]

3. **Substitute Known Values**:
We need to substitute the known values into the equation:
- Planck's constant, \( h = 6.626 \times 10^{-34} \) J·s (approximately \( 6.6 \times 10^{-34} \) J·s),
- Speed of light, \( c = 3 \times 10^8 \) m/s,
- Work function, \( \phi = 1.8 \) eV. We need to convert this to joules:
\[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \implies \phi = 1.8 \times 1.6 \times 10^{-19} \text{ J} = 2.88 \times 10^{-19} \text{ J} \]

4. **Calculate the Threshold Wavelength**:
Now we can substitute these values into the rearranged equation:
\[ \lambda_0 = \frac{(6.626 \times 10^{-34} \text{ J·s})(3 \times 10^8 \text{ m/s})}{2.88 \times 10^{-19} \text{ J}} \]

5. **Perform the Calculation**:
Calculate the numerator:
\[ 6.626 \times 10^{-34} \times 3 \times 10^8 = 1.9878 \times 10^{-25} \text{ J·m} \]
Now divide by the work function in joules:
\[ \lambda_0 = \frac{1.9878 \times 10^{-25}}{2.88 \times 10^{-19}} \approx 6.9 \times 10^{-7} \text{ m} \]

6. **Convert to Nanometers**:
Convert meters to nanometers (1 m = \( 10^9 \) nm):
\[ \lambda_0 \approx 689 \text{ nm} \]

### Final Answer:
The threshold wavelength for the cesium surface is approximately **689 nm**.

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