The peak emission from a black body at a certain temprature occurs at a wavelength of `9000 Å`. On increase its temperature , the total radiation emmited is increased its `81` times. At the intial temperature when the peak radiation from the black body is incident on a metal surface , it does not cause any photoemission from the surface . After the increase of temperature, the peak from the black body caused photoemission. To bring these photoelectrons to rest , a potential equivalent to the excitation energy between `n = 2 and n = 3` bohr levels of hydrogen atoms is required. Find the work function of the metal.

Explanation

To solve the problem step by step, we will break down the information given and apply relevant physics principles.

### Step 1: Understand the relationship between temperature and radiation
According to Stefan-Boltzmann law, the total radiative power (σ) emitted by a black body is proportional to the fourth power of its temperature (T):
\[ \sigma \propto T^4 \]

### Step 2: Relate the increase in total radiation to temperature
We are given that the total radiation emitted increases by a factor of 81. Thus, we can write:
\[ \frac{\sigma_2}{\sigma_1} = 81 \]
Using the relationship from Stefan-Boltzmann law:
\[ \frac{T_2^4}{T_1^4} = 81 \]
Taking the fourth root of both sides:
\[ \frac{T_2}{T_1} = 3 \]
This implies:
\[ T_2 = 3T_1 \]

### Step 3: Use Wien's displacement law
Wien's displacement law states that the product of the peak wavelength (λ) and the temperature (T) is a constant:
\[ \lambda m \cdot T = \text{constant} \]
Thus, we can relate the initial and final states:
\[ \lambda_1 T_1 = \lambda_2 T_2 \]
Substituting \( T_2 = 3T_1 \):
\[ \lambda_1 T_1 = \lambda_2 (3T_1) \]
This simplifies to:
\[ \lambda_2 = \frac{\lambda_1}{3} \]

### Step 4: Calculate the new wavelength
Given that \( \lambda_1 = 9000 \, \text{Å} \):
\[ \lambda_2 = \frac{9000 \, \text{Å}}{3} = 3000 \, \text{Å} \]

### Step 5: Determine the energy of the photon corresponding to λ2
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h = 4.135667696 \times 10^{-15} \, \text{eV·s} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \)

Substituting the values:
\[ E = \frac{(4.135667696 \times 10^{-15} \, \text{eV·s})(3 \times 10^8 \, \text{m/s})}{3000 \times 10^{-10} \, \text{m}} \]
Calculating this gives:
\[ E = \frac{(1.240 \times 10^{-6} \, \text{eV·m})}{3 \times 10^{-7} \, \text{m}} = 4.133 \, \text{eV} \]

### Step 6: Relate the maximum kinetic energy to the work function
According to the photoelectric effect:
\[ K.E. = E - \phi \]
Where \( K.E. \) is the maximum kinetic energy of the emitted electrons, \( E \) is the energy of the incoming photon, and \( \phi \) is the work function of the metal.

### Step 7: Calculate the maximum kinetic energy
The potential needed to stop the photoelectrons is given as equivalent to the energy difference between the n=2 and n=3 levels of hydrogen:
\[ E = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Calculating this:
\[ E = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \left( \frac{9 - 4}{36} \right) = 13.6 \left( \frac{5}{36} \right) \]
\[ E = \frac{68}{36} \approx 1.89 \, \text{eV} \]

### Step 8: Solve for the work function
Now substituting into the equation:
\[ 1.89 = 4.133 - \phi \]
Rearranging gives:
\[ \phi = 4.133 - 1.89 \]
\[ \phi \approx 2.243 \, \text{eV} \]

### Final Answer
The work function of the metal is approximately **2.25 eV**.

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