The work function of a metal is 2.5 eV. Light of wavelength 3600 `Å` is incident on this metal surface. The velocity of emitted photoelectrons will be

Options

A. `3.8xx10^6m//s`
B. `5.2 xx10^4 m//s`
C. `4xx10^4m//s`
D. `6xx10^5 m//s`

Answer

To solve the problem, we will use the principles of the photoelectric effect. The steps are as follows:

### Step 1: Convert Wavelength to Energy
First, we need to calculate the energy of the incident photons using the wavelength given. The energy of a photon can be calculated using the formula:

\[ E = \frac{hc}{\lambda} \]

Where:
- \(E\) is the energy of the photon,
- \(h\) 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 in meters.

Given that the wavelength is \(3600 \, \text{Å}\) (which is \(3600 \times 10^{-10} \, \text{m}\)), we can substitute the values into the equation.

### Step 2: Calculate the Energy of the Photon
Substituting the values:

\[ E = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{3600 \times 10^{-10} \, \text{m}} \]

Calculating this gives:

\[ E = \frac{1.9878 \times 10^{-25}}{3.6 \times 10^{-7}} \approx 5.52 \times 10^{-19} \, \text{J} \]

### Step 3: Convert Energy from Joules to Electron Volts
To convert the energy from Joules to electron volts (1 eV = \(1.6 \times 10^{-19} \, \text{J}\)):

\[ E \approx \frac{5.52 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.45 \, \text{eV} \]

### Step 4: Calculate the Maximum Kinetic Energy of the Emitted Electrons
Using the photoelectric effect equation:

\[ KE_{max} = E - \phi \]

Where \(\phi\) is the work function of the metal (given as \(2.5 \, \text{eV}\)):

\[ KE_{max} = 3.45 \, \text{eV} - 2.5 \, \text{eV} = 0.95 \, \text{eV} \]

### Step 5: Convert Kinetic Energy to Joules
Now, we convert the kinetic energy back to Joules:

\[ KE_{max} = 0.95 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} \approx 1.52 \times 10^{-19} \, \text{J} \]

### Step 6: Use Kinetic Energy to Find Velocity
The kinetic energy of the emitted electrons can also be expressed as:

\[ KE_{max} = \frac{1}{2} mv^2 \]

Where \(m\) is the mass of an electron (\(9.11 \times 10^{-31} \, \text{kg}\)). Rearranging for \(v\):

\[ v = \sqrt{\frac{2 \times KE_{max}}{m}} \]

Substituting the values:

\[ v = \sqrt{\frac{2 \times 1.52 \times 10^{-19}}{9.11 \times 10^{-31}}} \]

Calculating this gives:

\[ v = \sqrt{\frac{3.04 \times 10^{-19}}{9.11 \times 10^{-31}}} \approx \sqrt{3.34 \times 10^{11}} \approx 5.79 \times 10^5 \, \text{m/s} \]

### Final Answer
The velocity of the emitted photoelectrons is approximately:

\[ v \approx 5.79 \times 10^5 \, \text{m/s} \]