If a surface has a work function 4.0 eV, what is the maximum velocity of electrons liberated from the surface when it is irradiated with ultraviolet radiation of wavelength 0.2`mum`?

Options

A. `4.4xx10^(5)ms^(-1)`
B. `8.8xx10^(7)ms^(-1)`
C. `8.8xx10^(5)ms^(-1)`
D. `4.4xx10^(7)ms^(-1)`

Explanation

To solve the problem step by step, we will use the concepts from the photoelectric effect and the relevant equations.

### Step 1: Understand the Problem
We are given:
- Work function (φ) = 4.0 eV
- Wavelength of ultraviolet radiation (λ) = 0.2 µm = 200 nm

We need to find the maximum velocity (v_max) of electrons liberated from the surface.

### Step 2: Calculate the Energy of the Photon
The energy of a photon (E) can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) (Planck's constant) = \( 4.1357 \times 10^{-15} \) eV·s
- \( c \) (speed of light) = \( 3 \times 10^8 \) m/s
- \( \lambda \) (wavelength) = \( 200 \times 10^{-9} \) m

Substituting the values:
\[ E = \frac{(4.1357 \times 10^{-15} \text{ eV·s})(3 \times 10^8 \text{ m/s})}{200 \times 10^{-9} \text{ m}} \]

Calculating this gives:
\[ E \approx 6.21 \text{ eV} \]

### Step 3: Calculate the Maximum Kinetic Energy of the Electrons
Using the photoelectric equation:
\[ K_{max} = E - \phi \]
where:
- \( K_{max} \) is the maximum kinetic energy of the electrons.
- \( E \) is the energy of the photon.
- \( \phi \) is the work function.

Substituting the values:
\[ K_{max} = 6.21 \text{ eV} - 4.0 \text{ eV} = 2.21 \text{ eV} \]

### Step 4: Convert Kinetic Energy to Joules
To find the maximum velocity, we need to convert the kinetic energy from eV to Joules. The conversion factor is:
\[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \]

Thus,
\[ K_{max} = 2.21 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} \]
\[ K_{max} \approx 3.536 \times 10^{-19} \text{ J} \]

### Step 5: Use the Kinetic Energy Formula to Find Maximum Velocity
The kinetic energy of an electron can also be expressed as:
\[ K_{max} = \frac{1}{2} mv_{max}^2 \]
where:
- \( m \) is the mass of the electron = \( 9.11 \times 10^{-31} \) kg.

Rearranging the formula to solve for \( v_{max} \):
\[ v_{max} = \sqrt{\frac{2K_{max}}{m}} \]

Substituting the values:
\[ v_{max} = \sqrt{\frac{2 \times 3.536 \times 10^{-19}}{9.11 \times 10^{-31}}} \]

Calculating this gives:
\[ v_{max} \approx 8.84 \times 10^5 \text{ m/s} \]

### Final Answer
The maximum velocity of electrons liberated from the surface is approximately:
\[ v_{max} \approx 8.84 \times 10^5 \text{ m/s} \]

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