A photon of wavelength `4 xx 10^(-7)m` strikes on metal surface, The work function of the metal is `2.13eV`. The velocity of the photo electron is

Options

A. `5.67 xx 10^(6)ms^(-1)`
B. `5.67 xx 10^(5)ms^(-1)`
C. `5.67 xx 10^(-5)ms^(-1)`
D. `5.67 xx 10^(-6)ms^(-1)`

Explanation

To find the velocity of the photoelectron emitted from a metal surface when a photon strikes it, we can use the principles of the photoelectric effect as described by Einstein. Here’s a step-by-step solution:

### Step 1: Calculate the energy of the photon
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) 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 of the photon (\( 4 \times 10^{-7} \, \text{m} \))

Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{4 \times 10^{-7} \, \text{m}} \]
Calculating this gives:
\[ E = \frac{1.9878 \times 10^{-25} \, \text{J m}}{4 \times 10^{-7} \, \text{m}} = 4.9695 \times 10^{-19} \, \text{J} \]

### Step 2: Convert the work function from eV to Joules
The work function (\( \phi \)) is given as \( 2.13 \, \text{eV} \). To convert this to Joules, we use the conversion factor \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \):
\[ \phi = 2.13 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = 3.4122 \times 10^{-19} \, \text{J} \]

### Step 3: Calculate the kinetic energy of the emitted electron
According to the photoelectric effect, the kinetic energy (\( KE \)) of the emitted electron can be calculated using the equation:
\[ KE = E - \phi \]
Substituting the values:
\[ KE = 4.9695 \times 10^{-19} \, \text{J} - 3.4122 \times 10^{-19} \, \text{J} = 1.5573 \times 10^{-19} \, \text{J} \]

### Step 4: Relate kinetic energy to velocity
The kinetic energy of the electron is also given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the electron (\( 9.1 \times 10^{-31} \, \text{kg} \)). Rearranging this formula to solve for \( v \):
\[ v = \sqrt{\frac{2 \times KE}{m}} \]
Substituting the values:
\[ v = \sqrt{\frac{2 \times 1.5573 \times 10^{-19} \, \text{J}}{9.1 \times 10^{-31} \, \text{kg}}} \]
Calculating this gives:
\[ v = \sqrt{\frac{3.1146 \times 10^{-19}}{9.1 \times 10^{-31}}} = \sqrt{3.42 \times 10^{11}} \approx 5.86 \times 10^5 \, \text{m/s} \]

### Final Answer
The velocity of the photoelectron is approximately:
\[ v \approx 5.86 \times 10^5 \, \text{m/s} \]

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