Energy required to remove an electron from aluminium surface is `4.2 eV`. If light of wavelength `2000 Å` falls on the surface , the velocity of the fastest electron ejected from the surface will be

Options

A. `8.4 xx 10^(5) m//sec`
B. `7.4 xx 10^(5) m//sec`
C. `6.4 xx 10^(5) m//sec`
D. `8.4 xx 10^(6) m//sec`

Explanation

To solve the problem, we need to find the velocity of the fastest electron ejected from the aluminum surface when light of wavelength 2000 Å falls on it. Here are the steps to arrive at the solution:

### Step 1: Determine the Work Function
The work function (Φ) is given as 4.2 eV. This is the minimum energy required to remove an electron from the aluminum surface.

### Step 2: Calculate the Energy of the Incident Light
The energy (E) of the incident light 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) = 2000 Å = \( 2000 \times 10^{-10} \) m = \( 2 \times 10^{-7} \) m

Substituting the values:
\[ E = \frac{(4.1357 \times 10^{-15} \text{ eV·s})(3 \times 10^8 \text{ m/s})}{2 \times 10^{-7} \text{ m}} \]

Calculating this gives:
\[ E = \frac{1.24071 \times 10^{-6} \text{ eV·m}}{2 \times 10^{-7} \text{ m}} = 6.20355 \text{ eV} \]

### Step 3: Calculate the Maximum Kinetic Energy of the Ejected Electrons
The maximum kinetic energy (K.E.) of the ejected electrons can be calculated using the formula:
\[ K.E. = E - \Phi \]
Substituting the values:
\[ K.E. = 6.20355 \text{ eV} - 4.2 \text{ eV} = 2.00355 \text{ eV} \]

### Step 4: Convert Kinetic Energy to Joules
To convert the kinetic energy from eV to Joules, we use the conversion factor \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \):
\[ K.E. = 2.00355 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 3.20568 \times 10^{-19} \text{ J} \]

### Step 5: Use the Kinetic Energy to Find the Velocity
The kinetic energy can also be expressed in terms of mass and velocity:
\[ K.E. = \frac{1}{2} mv^2 \]
Where \( m \) (mass of an electron) = \( 9.1 \times 10^{-31} \text{ kg} \).

Rearranging for \( v \):
\[ v = \sqrt{\frac{2 \times K.E.}{m}} \]
Substituting the values:
\[ v = \sqrt{\frac{2 \times 3.20568 \times 10^{-19} \text{ J}}{9.1 \times 10^{-31} \text{ kg}}} \]

Calculating this gives:
\[ v = \sqrt{\frac{6.41136 \times 10^{-19}}{9.1 \times 10^{-31}}} = \sqrt{7.043 \times 10^{11}} \approx 8.4 \times 10^5 \text{ m/s} \]

### Final Answer
The velocity of the fastest electron ejected from the surface is approximately \( 8.4 \times 10^5 \text{ m/s} \).

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