The work function of metal is `1 eV`. Light of wavelength `3000 Å` is incident on this metal surface . The velocity of emitted photo - electrons will be

Options

A. `10 m//sec`
B. `1 xx 10^(3) m//sec`
C. `1 xx 10^(4) m//sec`
D. `1 xx 10^(6) m//sec`

Explanation

To find the velocity of emitted photoelectrons when light of wavelength 3000 Å is incident on a metal surface with a work function of 1 eV, we can follow these steps:

### Step 1: Calculate the energy of the incident photons
The energy of a photon 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) = \( 3000 \) Å = \( 3000 \times 10^{-10} \) m = \( 3 \times 10^{-7} \) m

Using the formula:
\[ E = \frac{(4.1357 \times 10^{-15} \text{ eV·s})(3 \times 10^8 \text{ m/s})}{3 \times 10^{-7} \text{ m}} \]
Calculating this gives:
\[ E = \frac{1.24071 \times 10^{-6} \text{ eV·m}}{3 \times 10^{-7} \text{ m}} = 4.1357 \text{ eV} \]

### Step 2: Calculate the maximum kinetic energy of the emitted photoelectrons
The maximum kinetic energy (K.E.) of the emitted photoelectrons can be calculated using the equation:
\[ K.E. = E - \phi \]
where \( \phi \) is the work function of the metal. Given that \( \phi = 1 \text{ eV} \):
\[ K.E. = 4.1357 \text{ eV} - 1 \text{ eV} = 3.1357 \text{ eV} \]

### Step 3: Convert kinetic energy from eV to Joules
To convert the kinetic energy from electron volts to Joules, we use the conversion factor \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \):
\[ K.E. = 3.1357 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 5.01712 \times 10^{-19} \text{ J} \]

### Step 4: Use the kinetic energy to find the velocity of the emitted electrons
The kinetic energy is also given by the equation:
\[ K.E. = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the electron (\( 9.11 \times 10^{-31} \text{ kg} \)). Rearranging for \( v \):
\[ v = \sqrt{\frac{2 \times K.E.}{m}} = \sqrt{\frac{2 \times 5.01712 \times 10^{-19} \text{ J}}{9.11 \times 10^{-31} \text{ kg}}} \]
Calculating this gives:
\[ v = \sqrt{\frac{1.003424 \times 10^{-18}}{9.11 \times 10^{-31}}} = \sqrt{1.101 \times 10^{12}} \approx 1.05 \times 10^6 \text{ m/s} \]

### Final Answer
The velocity of the emitted photoelectrons is approximately:
\[ v \approx 1.05 \times 10^6 \text{ m/s} \]

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