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 (Φ) of the metal given the stopping potentials for two different wavelengths, we can use the photoelectric effect principles. Here's a step-by-step solution:

### Step 1: Understand the relationship between stopping potential and photon energy
The stopping potential (V₀) is related to the maximum kinetic energy (K.E.) of the emitted electrons. According to the photoelectric equation:
\[ K.E. = eV_0 = E_{photon} - \Phi \]
where:
- \( e \) is the charge of the electron,
- \( V_0 \) is the stopping potential,
- \( E_{photon} \) is the energy of the incoming photon,
- \( \Phi \) is the work function.

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

### Step 3: Set up equations for both wavelengths
For the first wavelength (\( \lambda_1 = 4000 \, \text{Å} \), \( V_0 = 1.3 \, \text{V} \)):
\[ e(1.3) + \Phi = \frac{hc}{4000 \times 10^{-10}} \]

For the second wavelength (\( \lambda_2 = 4500 \, \text{Å} \), \( V_0 = 0.9 \, \text{V} \)):
\[ e(0.9) + \Phi = \frac{hc}{4500 \times 10^{-10}} \]

### Step 4: Rearrange the equations
From the first equation:
\[ \Phi = \frac{hc}{4000 \times 10^{-10}} - e(1.3) \]

From the second equation:
\[ \Phi = \frac{hc}{4500 \times 10^{-10}} - e(0.9) \]

### Step 5: Set the equations equal to each other
Since both expressions equal Φ, we can set them equal:
\[ \frac{hc}{4000 \times 10^{-10}} - e(1.3) = \frac{hc}{4500 \times 10^{-10}} - e(0.9) \]

### Step 6: Solve for Φ
Rearranging gives:
\[ \frac{hc}{4000 \times 10^{-10}} - \frac{hc}{4500 \times 10^{-10}} = e(1.3) - e(0.9) \]

Factoring out \( hc \):
\[ hc \left( \frac{1}{4000} - \frac{1}{4500} \right) = e(1.3 - 0.9) \]

### Step 7: Calculate the left side
Calculate the difference:
\[ \frac{1}{4000} - \frac{1}{4500} = \frac{4500 - 4000}{4000 \times 4500} = \frac{500}{18000000} = \frac{1}{36000} \]

### Step 8: Substitute values for hc and e
Using \( h = 6.626 \times 10^{-34} \, \text{Js} \) and \( c = 3 \times 10^8 \, \text{m/s} \):
\[ hc = 6.626 \times 10^{-34} \times 3 \times 10^8 = 1.9878 \times 10^{-25} \, \text{Jm} \]

Now, converting \( e \) to Joules (1 eV = \( 1.6 \times 10^{-19} \, \text{J} \)):
\[ e(1.3 - 0.9) = e(0.4) = 0.4 \times 1.6 \times 10^{-19} = 6.4 \times 10^{-20} \, \text{J} \]

### Step 9: Solve for Φ
Substituting back:
\[ 1.9878 \times 10^{-25} \left( \frac{1}{36000} \right) = 6.4 \times 10^{-20} \]

Calculating gives:
\[ \Phi = 2.3 \, \text{eV} \]

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