When photons of energy `4.25 eV` strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy `T_(A)` eV and De-broglie wavelength `lambda_(A)`. The maximum energy of photoelectron liberated from another metal B by photon of energy 4.70 eV is `T_(B) = (T_(A) - 1.50) eV` if the de Brogle wavelength of these photoelectrons is `lambda_(B) = 2 lambda_(A) `, then

Options

A. the work function of A is 2.25 eV
B. the work funtion of B is 4.20 eV
C. `T_A=2.00eV`
D. `T_B=2.75eV`

Explanation

To solve the problem step by step, we will analyze the information provided and apply the relevant equations from the photoelectric effect and de Broglie wavelength.

### Step 1: Understand the problem
We have two metals A and B with the following information:
- For metal A:
- Photon energy \( E_A = 4.25 \, \text{eV} \)
- Maximum kinetic energy of photoelectrons \( T_A \)
- de Broglie wavelength \( \lambda_A \)

- For metal B:
- Photon energy \( E_B = 4.70 \, \text{eV} \)
- Maximum kinetic energy of photoelectrons \( T_B = T_A - 1.50 \, \text{eV} \)
- de Broglie wavelength \( \lambda_B = 2 \lambda_A \)

### Step 2: Relate kinetic energy and de Broglie wavelength
The kinetic energy of the photoelectrons can be related to their momentum and de Broglie wavelength. The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum. The kinetic energy \( T \) is related to momentum by:
\[ p = \sqrt{2mT} \]
Thus, we can express the kinetic energy in terms of the de Broglie wavelength:
\[ \lambda = \frac{h}{\sqrt{2mT}} \implies T = \frac{h^2}{2m\lambda^2} \]

### Step 3: Set up the equations for metals A and B
Using the above relation, we can write:
\[ T_A \propto \frac{1}{\lambda_A^2} \quad \text{and} \quad T_B \propto \frac{1}{\lambda_B^2} \]
Given that \( \lambda_B = 2 \lambda_A \):
\[ T_B \propto \frac{1}{(2\lambda_A)^2} = \frac{1}{4\lambda_A^2} \]

### Step 4: Relate \( T_A \) and \( T_B \)
From the problem, we know:
\[ \frac{T_A}{T_B} = \frac{\lambda_B^2}{\lambda_A^2} \]
Substituting \( \lambda_B = 2\lambda_A \):
\[ \frac{T_A}{T_B} = \frac{(2\lambda_A)^2}{\lambda_A^2} = 4 \]
Thus:
\[ T_A = 4 T_B \]

### Step 5: Substitute \( T_B \) in terms of \( T_A \)
From the relation \( T_B = T_A - 1.50 \):
\[ T_A = 4(T_A - 1.50) \]
Expanding this gives:
\[ T_A = 4T_A - 6 \]
Rearranging:
\[ 3T_A = 6 \implies T_A = 2 \, \text{eV} \]

### Step 6: Calculate \( T_B \)
Now substituting \( T_A \) back to find \( T_B \):
\[ T_B = T_A - 1.50 = 2 - 1.50 = 0.5 \, \text{eV} \]

### Step 7: Calculate work functions
Using the photoelectric equation:
\[ E = \phi + T \]
For metal A:
\[ 4.25 = \phi_A + 2 \implies \phi_A = 4.25 - 2 = 2.25 \, \text{eV} \]

For metal B:
\[ 4.70 = \phi_B + 0.5 \implies \phi_B = 4.70 - 0.5 = 4.20 \, \text{eV} \]

### Final Results
- Kinetic energy of photoelectrons from metal A: \( T_A = 2 \, \text{eV} \)
- Kinetic energy of photoelectrons from metal B: \( T_B = 0.5 \, \text{eV} \)
- Work function of metal A: \( \phi_A = 2.25 \, \text{eV} \)
- Work function of metal B: \( \phi_B = 4.20 \, \text{eV} \)