The work function for tungsten and sodium are `4.5 eV` and `2.3 eV` respectively . If the threshold wavelength `lambda` for sodium is `5460 Å`, the value of `lambda` for tungsten is

Options

A. `5893 A`
B. `10683 A`
C. `2791 A`
D. `528 A`

Explanation

To find the threshold wavelength for tungsten given the work functions and the threshold wavelength for sodium, we can follow these steps:

### Step 1: Understand the relationship between work function and threshold wavelength
The work function \( W \) and the threshold wavelength \( \lambda_0 \) are related by the equation:
\[ W = \frac{hc}{\lambda_0} \]
where:
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda_0 \) is the threshold wavelength.

### Step 2: Set up the equation for both materials
For sodium (1):
\[ W_1 = 2.3 \, \text{eV}, \quad \lambda_{01} = 5460 \, \text{Å} \]
For tungsten (2):
\[ W_2 = 4.5 \, \text{eV}, \quad \lambda_{02} = ? \]

Using the relationship, we can write:
\[ W_1 \lambda_{01} = W_2 \lambda_{02} \]

### Step 3: Substitute the known values into the equation
Substituting the values we have:
\[ (2.3 \, \text{eV}) \times (5460 \, \text{Å}) = (4.5 \, \text{eV}) \times \lambda_{02} \]

### Step 4: Solve for \( \lambda_{02} \)
Rearranging the equation to solve for \( \lambda_{02} \):
\[ \lambda_{02} = \frac{(2.3 \, \text{eV}) \times (5460 \, \text{Å})}{4.5 \, \text{eV}} \]

### Step 5: Calculate the value
Now, let's calculate:
\[ \lambda_{02} = \frac{(2.3 \times 5460)}{4.5} \]
Calculating the numerator:
\[ 2.3 \times 5460 = 12558 \]
Now divide by 4.5:
\[ \lambda_{02} = \frac{12558}{4.5} \approx 2791.33 \, \text{Å} \]

### Final Answer
Thus, the threshold wavelength \( \lambda_{02} \) for tungsten is approximately:
\[ \lambda_{02} \approx 2791 \, \text{Å} \]