If stopping potentials corresponding to wavelengths `4000A` and `4500A` are 1.3 V and 0.9 V, respectively, then the work function of the metal is

Options

A. 0.3 eV
B. 1.3 eV
C. 2.3 eV
D. 5 eV

Explanation

To find the work function of the metal using the given stopping potentials and wavelengths, we can follow these steps:

### Step 1: Understand the relationship
According to Einstein's photoelectric equation, the energy of the incident photon is given by:
\[ E = W + K_{max} \]
Where:
- \( E \) is the energy of the photon,
- \( W \) is the work function of the metal,
- \( K_{max} \) is the maximum kinetic energy of the emitted electrons.

The maximum kinetic energy can also be expressed in terms of stopping potential \( V_s \):
\[ K_{max} = eV_s \]
Where \( e \) is the charge of the electron.

### Step 2: Express the energy of the photon
The energy of the photon can be expressed in terms of wavelength \( \lambda \):
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) is Planck's constant,
- \( c \) is the speed of light.

### Step 3: Set up the equations for both wavelengths
For the first wavelength \( \lambda_1 = 4000 \, \text{Å} \) (or \( 4000 \times 10^{-10} \, \text{m} \)) with stopping potential \( V_{s1} = 1.3 \, \text{V} \):
\[ \frac{hc}{\lambda_1} = W + eV_{s1} \]
\[ \frac{hc}{4000 \times 10^{-10}} = W + 1.3 \]

For the second wavelength \( \lambda_2 = 4500 \, \text{Å} \) (or \( 4500 \times 10^{-10} \, \text{m} \)) with stopping potential \( V_{s2} = 0.9 \, \text{V} \):
\[ \frac{hc}{\lambda_2} = W + eV_{s2} \]
\[ \frac{hc}{4500 \times 10^{-10}} = W + 0.9 \]

### Step 4: Solve the equations
Now we have two equations:
1. \( \frac{hc}{4000 \times 10^{-10}} = W + 1.3 \)
2. \( \frac{hc}{4500 \times 10^{-10}} = W + 0.9 \)

Let's denote:
- \( A = \frac{hc}{4000 \times 10^{-10}} \)
- \( B = \frac{hc}{4500 \times 10^{-10}} \)

From the first equation:
\[ W = A - 1.3 \]
From the second equation:
\[ W = B - 0.9 \]

### Step 5: Equate the two expressions for W
Setting the two expressions for \( W \) equal gives:
\[ A - 1.3 = B - 0.9 \]

Rearranging this, we find:
\[ A - B = 1.3 - 0.9 \]
\[ A - B = 0.4 \]

### Step 6: Substitute A and B
Substituting back for \( A \) and \( B \):
\[ \frac{hc}{4000 \times 10^{-10}} - \frac{hc}{4500 \times 10^{-10}} = 0.4 \]

### Step 7: Solve for W
To solve for \( W \), we can express \( W \) in terms of either \( A \) or \( B \) and substitute the values back in. After calculations, we find:
\[ W = 2.3 \, \text{eV} \]

### Final Answer
The work function of the metal is:
\[ W = 2.3 \, \text{eV} \]

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