The work function of a metal is 3.4 eV. A light of wavelength `3000Å` is incident on it. The maximum kinetic energy of the ejected electron will be :

Options

A. `1.18 xx 10^(-19)J`
B. `2.36 xx 10^(-19)J`
C. `3.4 xx 10^(-19)J`
D. Electron will not be ejected

Answer

To solve the problem step by step, we will use the Einstein photoelectric equation, which relates the maximum kinetic energy of ejected electrons to the energy of the incident photons and the work function of the metal.

### Step 1: Understand the Problem
We are given:
- Work function (φ) of the metal = 3.4 eV
- Wavelength (λ) of the incident light = 3000 Å (angstroms)

We need to find the maximum kinetic energy (K_max) of the ejected electron.

### Step 2: Convert Work Function to Joules
The work function is given in electron volts (eV), and we need to convert it to joules (J) for our calculations.

1 eV = \(1.6 \times 10^{-19}\) J

So,
\[ \phi = 3.4 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 5.44 \times 10^{-19} \, \text{J} \]

### Step 3: Calculate the Energy of the Incident Photon
The energy (E) of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \(h\) = Planck's constant = \(6.626 \times 10^{-34} \, \text{J s}\)
- \(c\) = Speed of light = \(3 \times 10^8 \, \text{m/s}\)
- \(\lambda\) = Wavelength in meters = \(3000 \, \text{Å} = 3000 \times 10^{-10} \, \text{m}\)

Now substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{3000 \times 10^{-10} \, \text{m}} \]

Calculating this gives:
\[ E = \frac{1.9878 \times 10^{-25}}{3000 \times 10^{-10}} = 6.626 \times 10^{-19} \, \text{J} \]

### Step 4: Calculate Maximum Kinetic Energy
Using the Einstein photoelectric equation:
\[ K_{max} = E - \phi \]

Substituting the values we found:
\[ K_{max} = (6.626 \times 10^{-19} \, \text{J}) - (5.44 \times 10^{-19} \, \text{J}) \]
\[ K_{max} = 1.186 \times 10^{-19} \, \text{J} \]

### Step 5: Convert Kinetic Energy to eV
To convert the kinetic energy back to eV:
\[ K_{max} = \frac{1.186 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 0.74 \, \text{eV} \]

### Final Answer
The maximum kinetic energy of the ejected electron is approximately \(0.74 \, \text{eV}\).

---