What is the work function of metal (in eV) if stopping potential and saturation current for a light of wavelenth `6000 Å are 1.37 V and 0.40 muA` respectively?
`( " Take " h=4.14xx10^(-15) eVs)`

Explanation

To find the work function of the metal, we can use the Einstein photoelectric equation, which relates the energy of the incident photons to the work function and the maximum kinetic energy of the emitted electrons.

### Step 1: Calculate the energy of the incident photon (E)

The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]

Where:
- \( h \) is Planck's constant (\( 4.14 \times 10^{-15} \, \text{eV s} \))
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \))
- \( \lambda \) is the wavelength of the light in meters.

Given:
- \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \)

Now substituting the values:
\[ E = \frac{(4.14 \times 10^{-15} \, \text{eV s})(3 \times 10^8 \, \text{m/s})}{6 \times 10^{-7} \, \text{m}} \]

Calculating this gives:
\[ E = \frac{(4.14 \times 3) \times 10^{-7} \, \text{eV}}{6} \]
\[ E = \frac{12.42 \times 10^{-7}}{6} \]
\[ E = 2.07 \, \text{eV} \]

### Step 2: Use the stopping potential to find the work function

The stopping potential \( V_0 \) is given as \( 1.37 \, \text{V} \). According to the Einstein photoelectric equation:
\[ E = \phi + K_{\text{max}} \]
Where:
- \( \phi \) is the work function
- \( K_{\text{max}} \) can be expressed as \( eV_0 \)

Thus, we can rearrange the equation to find the work function:
\[ \phi = E - eV_0 \]

Substituting the values:
\[ \phi = 2.07 \, \text{eV} - 1.37 \, \text{V} \]
\[ \phi = 2.07 - 1.37 \]
\[ \phi = 0.70 \, \text{eV} \]

### Final Answer
The work function of the metal is \( 0.70 \, \text{eV} \).

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