what is the work fuction of the metal if the light of wavelength 4000Å generates photoelectrons of velocity `6xx10^(5) ms^(-1)` from it ?
(Mass o felectron `=9xx10^(-31)`kg
Velocity of light `=3xx10^(8)ms^(-1)`
Planck's constant `=6.626xx10^(-34)Js`
Charge of electron `=1.6xx10^(-19)jeV^(-1)`

Options

A. 2.1 eV
B. 3.1 eV
C. 0.9 eV
D. 4.0 eV

Explanation

To find the work function of the metal, we can use the relationship between the total energy of the incoming photons, the work function (φ), and the kinetic energy (KE) of the emitted photoelectrons. The equation is given by:

\[ E_{total} = \phi + KE \]

Where:
- \( E_{total} \) is the energy of the incoming photons.
- \( φ \) is the work function of the metal.
- \( KE \) is the kinetic energy of the emitted photoelectrons.

### Step 1: Calculate the energy of the incoming photons

The energy of the incoming photons can be calculated using the formula:

\[ E_{total} = \frac{hc}{\lambda} \]

Where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \))
- \( c \) is the speed of light (\( 3 \times 10^{8} \, \text{m/s} \))
- \( \lambda \) is the wavelength of the light (4000 Å = \( 4000 \times 10^{-10} \, \text{m} \))

Substituting the values:

\[ E_{total} = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^{8} \, \text{m/s})}{4000 \times 10^{-10} \, \text{m}} \]

Calculating this gives:

\[ E_{total} = \frac{1.9878 \times 10^{-25} \, \text{J}}{4 \times 10^{-7}} = 4.9695 \times 10^{-19} \, \text{J} \]

### Step 2: Calculate the kinetic energy of the emitted photoelectrons

The kinetic energy (KE) of the emitted photoelectrons can be calculated using the formula:

\[ KE = \frac{1}{2} mv^2 \]

Where:
- \( m \) is the mass of the electron (\( 9 \times 10^{-31} \, \text{kg} \))
- \( v \) is the velocity of the emitted photoelectrons (\( 6 \times 10^{5} \, \text{m/s} \))

Substituting the values:

\[ KE = \frac{1}{2} (9 \times 10^{-31} \, \text{kg}) (6 \times 10^{5} \, \text{m/s})^2 \]

Calculating this gives:

\[ KE = \frac{1}{2} (9 \times 10^{-31}) (3.6 \times 10^{11}) = 1.62 \times 10^{-19} \, \text{J} \]

### Step 3: Calculate the work function

Now, we can find the work function (φ) using the relationship:

\[ \phi = E_{total} - KE \]

Substituting the values we calculated:

\[ \phi = (4.9695 \times 10^{-19} \, \text{J}) - (1.62 \times 10^{-19} \, \text{J}) \]

Calculating this gives:

\[ \phi = 3.3495 \times 10^{-19} \, \text{J} \]

### Step 4: Convert the work function to electron volts

To convert Joules to electron volts, we use the conversion factor:

\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]

Thus, we can convert φ:

\[ \phi = \frac{3.3495 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \]

Calculating this gives:

\[ \phi \approx 2.093 \, \text{eV} \]

### Final Answer

The work function of the metal is approximately \( 2.1 \, \text{eV} \).

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