Light of wavelength 4000 Å is incident on a sodium surface for which the threshold wavelength of photoelectrons is 5420 Å. The work function of sodium is

Options

A. 0.57 eV
B. 1.14 eV
C. 2.29 eV
D. 4.58 eV

Explanation

To find the work function of sodium when light of wavelength 4000 Å is incident on its surface, we can follow these steps:

### Step 1: Understand the relationship between wavelength and frequency
The frequency (ν) of light is related to its wavelength (λ) by the equation:
\[ c = ν \cdot λ \]
where \( c \) is the speed of light (\( 3 \times 10^8 \) m/s).

### Step 2: Convert the given wavelengths from Ångströms to meters
The given wavelengths are:
- Incident light wavelength: \( 4000 \, \text{Å} = 4000 \times 10^{-10} \, \text{m} \)
- Threshold wavelength: \( 5420 \, \text{Å} = 5420 \times 10^{-10} \, \text{m} \)

### Step 3: Calculate the threshold frequency (ν₀)
Using the threshold wavelength (λ₀):
\[ ν₀ = \frac{c}{λ₀} \]
Substituting the values:
\[ ν₀ = \frac{3 \times 10^8 \, \text{m/s}}{5420 \times 10^{-10} \, \text{m}} \]

### Step 4: Calculate the work function (φ)
The work function can be calculated using the equation:
\[ φ = hν₀ \]
where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)).

### Step 5: Convert the work function from Joules to electron volts
To convert Joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ φ \, (\text{in eV}) = \frac{φ \, (\text{in J})}{1.6 \times 10^{-19}} \]

### Step 6: Substitute and calculate
1. Calculate \( ν₀ \):
\[ ν₀ = \frac{3 \times 10^8}{5420 \times 10^{-10}} \approx 5.53 \times 10^{14} \, \text{Hz} \]

2. Calculate the work function \( φ \):
\[ φ = hν₀ = 6.626 \times 10^{-34} \times 5.53 \times 10^{14} \approx 3.67 \times 10^{-19} \, \text{J} \]

3. Convert to electron volts:
\[ φ \, (\text{in eV}) = \frac{3.67 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.29 \, \text{eV} \]

### Final Answer
The work function of sodium is approximately **2.29 eV**.

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