To solve the problem step by step, we will follow the reasoning laid out in the video transcript.
### Step 1: Understanding the relationship between temperatures and radiation
According to Stefan's law, the radiant heat energy emitted from a unit area in one second is given by:
\[ E = \sigma T^4 \]
where \( E \) is the radiant energy, \( \sigma \) is the Stefan-Boltzmann constant, and \( T \) is the absolute temperature.
Let:
- \( E_1 = \sigma T_1^4 \) (at temperature \( T_1 \))
- \( E_2 = \sigma T_2^4 \) (at temperature \( T_2 \))
Given that the total radiation increases 16 times when the temperature changes from \( T_1 \) to \( T_2 \), we have:
\[ E_2 = 16 E_1 \]
### Step 2: Setting up the equations
Substituting the expressions for \( E_1 \) and \( E_2 \):
\[ \sigma T_2^4 = 16 \sigma T_1^4 \]
We can cancel \( \sigma \) from both sides:
\[ T_2^4 = 16 T_1^4 \]
Taking the fourth root of both sides:
\[ T_2 = 2 T_1 \]
### Step 3: Using Wien's Law to find the new wavelength
Wien's Law states:
\[ \lambda_1 T_1 = \lambda_2 T_2 \]
Given that \( \lambda_1 = 9000 \, \text{Å} \) and substituting \( T_2 = 2 T_1 \):
\[ 9000 \, \text{Å} \cdot T_1 = \lambda_2 \cdot (2 T_1) \]
Cancelling \( T_1 \) from both sides:
\[ \lambda_2 = \frac{9000 \, \text{Å}}{2} = 4500 \, \text{Å} \]
### Step 4: Finding the maximum kinetic energy of photoelectrons
The maximum kinetic energy of the photoelectrons is equal to the energy of the photon emitted when an electron transitions from the M-shell (n=3) to the L-shell (n=2) in a hydrogen atom.
The energy difference is given by:
\[ KE = -13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( n_1 = 2 \) and \( n_2 = 3 \):
\[ KE = -13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
\[ KE = -13.6 \left( \frac{1}{4} - \frac{1}{9} \right) \]
Calculating the fractions:
\[ KE = -13.6 \left( \frac{9 - 4}{36} \right) = -13.6 \left( \frac{5}{36} \right) \]
\[ KE = -13.6 \times 0.1389 \approx 1.89 \, \text{eV} \]
### Step 5: Finding the work function of the metal
The work function \( \phi \) of the metal can be found using the equation:
\[ \phi = E - KE \]
where \( E \) is the energy of the photon corresponding to the wavelength \( \lambda_2 \):
\[ E = \frac{hc}{\lambda_2} \]
Substituting \( h = 6.626 \times 10^{-34} \, \text{Js} \) and \( c = 3 \times 10^8 \, \text{m/s} \):
Convert \( \lambda_2 = 4500 \, \text{Å} = 4500 \times 10^{-10} \, \text{m} \):
\[ E = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{4500 \times 10^{-10}} \]
Calculating:
\[ E \approx 2.75 \, \text{eV} \]
Now, substituting back to find the work function:
\[ \phi = 2.75 \, \text{eV} - 1.89 \, \text{eV} = 0.86 \, \text{eV} \]
### Final Answer
The work function of the metal is approximately:
\[ \phi \approx 0.88 \, \text{eV} \]
