Ejection of the photoelectron from metal in the photoelectric effect experiment can be stopped by applying 0.5 V when the radiation of 250 nm is used. The work function of the metal is

Options

A. 5 eV
B. 4 eV
C. 5.5 eV
D. 4.5 eV

Explanation

To find the work function (φ) of the metal in the photoelectric effect experiment, we can use the following steps:

### Step 1: Understand the relationship between energy, work function, and kinetic energy
The energy of the incoming photon (E) can be expressed as:
\[ E = \phi + KE \]
Where:
- \( E \) is the energy of the photon,
- \( \phi \) is the work function of the metal,
- \( KE \) is the kinetic energy of the ejected electron.

### Step 2: Calculate the energy of the photon
The energy of the photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength of the radiation (in meters).

Given that the wavelength \( \lambda = 250 \, \text{nm} = 250 \times 10^{-9} \, \text{m} \), we can substitute the values into the equation.

### Step 3: Substitute the values into the energy equation
Substituting the values into the energy equation:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{250 \times 10^{-9} \, \text{m}} \]

### Step 4: Calculate the energy in joules
Calculating this gives:
\[ E = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{250 \times 10^{-9}} \]
\[ E = \frac{1.9878 \times 10^{-25}}{250 \times 10^{-9}} \]
\[ E = 7.9512 \times 10^{-19} \, \text{J} \]

### Step 5: Convert energy from joules to electron volts
To convert joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E = \frac{7.9512 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 4.969 \, \text{eV} \]

### Step 6: Use the stopping potential to find kinetic energy
The stopping potential \( V_s \) is given as 0.5 V, which is equivalent to 0.5 eV. The kinetic energy of the ejected electron can be expressed as:
\[ KE = eV_s = 0.5 \, \text{eV} \]

### Step 7: Substitute into the energy equation
Now, substituting the values into the energy equation:
\[ 4.969 \, \text{eV} = \phi + 0.5 \, \text{eV} \]

### Step 8: Solve for the work function
Rearranging the equation to solve for φ:
\[ \phi = 4.969 \, \text{eV} - 0.5 \, \text{eV} \]
\[ \phi = 4.469 \, \text{eV} \]

### Step 9: Round to the appropriate significant figures
Rounding this to one decimal place gives:
\[ \phi \approx 4.5 \, \text{eV} \]

### Final Answer
The work function of the metal is approximately **4.5 eV**.

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