If the threshold wavelength for sodium is `5420 Å`, then the work function of sodium is

Options

A. `4.58 eV`
B. `2.28 eV`
C. `1.14 eV`
D. `0.23 eV`

Explanation

To find the work function (φ) of sodium given its threshold wavelength (λ₀) of 5420 Å, we can use the formula that relates the work function to the wavelength:

\[ \phi = \frac{hc}{\lambda_0} \]

Where:
- \( \phi \) is the work function in electron volts (eV),
- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \( c \) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \( \lambda_0 \) is the threshold wavelength in meters.

### Step-by-Step Solution:

1. **Convert the wavelength from angstroms to meters:**
\[ \lambda_0 = 5420 \, \text{Å} = 5420 \times 10^{-10} \, \text{m} = 5.42 \times 10^{-7} \, \text{m} \]

2. **Substitute the values into the work function formula:**
\[ \phi = \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{5.42 \times 10^{-7} \, \text{m}} \]

3. **Calculate the numerator:**
\[ hc = 6.626 \times 10^{-34} \times 3 \times 10^8 = 1.9878 \times 10^{-25} \, \text{Jm} \]

4. **Calculate the work function in joules:**
\[ \phi = \frac{1.9878 \times 10^{-25}}{5.42 \times 10^{-7}} = 3.66 \times 10^{-19} \, \text{J} \]

5. **Convert the work function from joules to electron volts:**
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
\[ \phi = \frac{3.66 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.29 \, \text{eV} \]

### Final Answer:
The work function of sodium is approximately \( \phi \approx 2.29 \, \text{eV} \).

---