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 amitted 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 photoelectrons emitted is

Options

A. 2 eV
B. 4 eV
C. 14 eV
D. 18 eV

Explanation

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

### Step 1: Understand the relationship between kinetic energy and work function
We know that the maximum kinetic energy (K.E.) of the emitted electrons can be expressed as:
\[ K.E. = E - \phi \]
where \( E \) is the energy of the incident photons and \( \phi \) is the work function of the metallic surface.

### Step 2: Determine the work function
From the problem, we are given that the maximum kinetic energy of the emitted electrons when light of wavelength corresponding to temperature \( T_1 \) is used is 6 eV, and this is three times the work function:
\[ K.E. = 3\phi \]
Thus, we can express the work function as:
\[ \phi = \frac{K.E.}{3} = \frac{6 \text{ eV}}{3} = 2 \text{ eV} \]

### Step 3: Calculate the energy of the incident photons at temperature \( T_1 \)
The energy of the photons can be calculated using the equation:
\[ E = K.E. + \phi \]
Substituting the known values:
\[ E = 6 \text{ eV} + 2 \text{ eV} = 8 \text{ eV} \]

### Step 4: Relate the energy of the photons to wavelength
The energy of the photons can also be expressed in terms of wavelength:
\[ E = \frac{hc}{\lambda} \]
Setting this equal to the energy calculated:
\[ \frac{hc}{\lambda_1} = 8 \text{ eV} \]

### Step 5: Analyze the situation at temperature \( T_2 \)
We are given that \( T_2 = 2T_1 \). The wavelength of the maximum intensity emitted from a black body is given by Wien's displacement law:
\[ \lambda_{max} \propto \frac{1}{T} \]
Thus, if \( T_2 = 2T_1 \), then:
\[ \lambda_2 = \frac{\lambda_1}{2} \]

### Step 6: Calculate the energy of the photons at temperature \( T_2 \)
Using the relationship for energy again, we find:
\[ E_2 = \frac{hc}{\lambda_2} = \frac{hc}{\frac{\lambda_1}{2}} = 2 \cdot \frac{hc}{\lambda_1} = 2E_1 \]
Since \( E_1 = 8 \text{ eV} \):
\[ E_2 = 2 \cdot 8 \text{ eV} = 16 \text{ eV} \]

### Step 7: Calculate the maximum kinetic energy at temperature \( T_2 \)
Now we can find the new maximum kinetic energy using the same formula:
\[ K.E. = E_2 - \phi \]
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:
\[ \boxed{14 \text{ eV}} \]