Ligth of wavelength `4000 Å` is incident on a metal plate whose work function is `2 eV`. What is maximum kinetic enegy of emitted photoelectron ?

Options

A. `0.5 eV`
B. `1.1 eV`
C. `2.0 eV`
D. `1.5 eV`

Explanation

To find the maximum kinetic energy of the emitted photoelectron when light of wavelength \(4000 \, \text{Å}\) is incident on a metal plate with a work function of \(2 \, \text{eV}\), we can follow these steps:

### Step 1: Convert Wavelength to Meters
The wavelength is given in angstroms. We need to convert it to meters for our calculations.
\[ \lambda = 4000 \, \text{Å} = 4000 \times 10^{-10} \, \text{m} = 4 \times 10^{-7} \, \text{m} \]

### Step 2: Calculate the Frequency of the Incident Light
The frequency \(f\) can be calculated using the formula:
\[ f = \frac{c}{\lambda} \]
where \(c\) is the speed of light, approximately \(3 \times 10^8 \, \text{m/s}\).
\[ f = \frac{3 \times 10^8 \, \text{m/s}}{4 \times 10^{-7} \, \text{m}} = 7.5 \times 10^{14} \, \text{Hz} \]

### Step 3: Calculate the Energy of the Incident Photons
The energy \(E\) of the photons can be calculated using the formula:
\[ E = h \cdot f \]
where \(h\) is Planck's constant, approximately \(6.63 \times 10^{-34} \, \text{Js}\).
\[ E = 6.63 \times 10^{-34} \, \text{Js} \cdot 7.5 \times 10^{14} \, \text{Hz} = 4.9725 \times 10^{-19} \, \text{J} \]

### Step 4: Convert Energy from Joules to Electron Volts
To convert the energy from joules to electron volts, we use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\).
\[ E = \frac{4.9725 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 3.105 \, \text{eV} \]

### Step 5: Calculate the Maximum Kinetic Energy of the Photoelectrons
The maximum kinetic energy \(K.E.\) of the emitted photoelectrons can be calculated using the formula:
\[ K.E. = E - \phi \]
where \(\phi\) is the work function of the metal, given as \(2 \, \text{eV}\).
\[ K.E. = 3.105 \, \text{eV} - 2 \, \text{eV} = 1.105 \, \text{eV} \]

### Final Answer
The maximum kinetic energy of the emitted photoelectron is approximately \(1.1 \, \text{eV}\).

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