If light of wavelength of maximum intensity emitted from surface at temperature `T_(1)` Is used to cause photoelectric emission from a metallic surface, the maximum kinetic energy of the emitted electron is 6 ev, which is 3 time the work function of the metallic surface. If light of wavelength of maximum intensity emitted from a surface at temperature `T_(2) (T_(2)=2T_(1))` is used, the maximum kinetic energy of the photo electrons emitted is

Options

A. 10eV
B. 8eV
C. 14eV
D. 12eV

Answer

To solve the problem step by step, we will analyze the given information and apply relevant physics concepts.

### Step 1: Understand the Given Information
We know that:
- The maximum kinetic energy (K.E.) of the emitted electrons when light of wavelength corresponding to temperature \( T_1 \) is used is \( 6 \, \text{eV} \).
- This kinetic energy is three times the work function (\( \phi \)) of the metallic surface.

From this, we can express the work function:
\[ K.E. = 6 \, \text{eV} = 3\phi \implies \phi = \frac{6 \, \text{eV}}{3} = 2 \, \text{eV} \]

### Step 2: Calculate Energy of the Incident Light at Temperature \( T_1 \)
Using the photoelectric equation:
\[ K.E. = \frac{hc}{\lambda_1} - \phi \]
Substituting the known values:
\[ 6 \, \text{eV} = \frac{hc}{\lambda_1} - 2 \, \text{eV} \]
This implies:
\[ \frac{hc}{\lambda_1} = 6 \, \text{eV} + 2 \, \text{eV} = 8 \, \text{eV} \]

### Step 3: Apply Wien's Displacement Law
According to Wien's displacement law, the wavelength of maximum intensity emitted from a black body is inversely proportional to its temperature. Thus, if \( T_2 = 2T_1 \), we have:
\[ \lambda_2 = \frac{\lambda_1}{2} \]

### Step 4: Calculate Energy of the Incident Light at Temperature \( T_2 \)
Now, we can find the energy of the light corresponding to \( T_2 \):
\[ \frac{hc}{\lambda_2} = \frac{hc}{\frac{\lambda_1}{2}} = 2 \cdot \frac{hc}{\lambda_1} \]
Substituting the value we found earlier:
\[ \frac{hc}{\lambda_2} = 2 \cdot 8 \, \text{eV} = 16 \, \text{eV} \]

### Step 5: Calculate the Maximum Kinetic Energy of Emitted Electrons at \( T_2 \)
Using the photoelectric equation again for \( T_2 \):
\[ K.E. = \frac{hc}{\lambda_2} - \phi \]
Substituting the values:
\[ K.E. = 16 \, \text{eV} - 2 \, \text{eV} = 14 \, \text{eV} \]

### Final Answer
The maximum kinetic energy of the photoelectrons emitted when light of wavelength corresponding to temperature \( T_2 \) is used is:
\[ \boxed{14 \, \text{eV}} \]