When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy, `T_A` (expressed in eV) and deBroglie wavelength `lambda_A`. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.20V is `T_B = T_A -1.50eV`. If the deBroglie wavelength of those photoelectrons is `lambda_B = 2lambda_A` then

Options

A. the work function of A is 2.25 eV
B. the work function of B is 3.70 eV
C. `T_(A)=2.00 eV`
D. `T_(B)=2.75 eV`

Explanation

To solve the problem step by step, we will use the concepts of photoelectric effect and de Broglie wavelength.

### Step 1: Write the equations for the work functions of metals A and B.
The work function (φ) of a metal can be calculated using the formula:
\[ \phi_A = E - T_A \]
\[ \phi_B = E - T_B \]
where \( E \) is the energy of the incoming photons.

For Metal A:
\[ \phi_A = 4.25 \, \text{eV} - T_A \] (Equation 1)

For Metal B:
\[ \phi_B = 4.20 \, \text{eV} - T_B \] (Equation 2)

### Step 2: Express \( T_B \) in terms of \( T_A \).
According to the problem, we have:
\[ T_B = T_A - 1.50 \, \text{eV} \] (Equation 3)

### Step 3: Use the de Broglie wavelength relationship.
The de Broglie wavelength (λ) of a particle is given by:
\[ \lambda = \frac{h}{\sqrt{2mT}} \]
where \( h \) is Planck's constant, \( m \) is the mass of the electron, and \( T \) is the kinetic energy.

We know that:
\[ \lambda_B = 2\lambda_A \]

Using the de Broglie wavelength formula for both metals, we can write:
\[ \lambda_B = \frac{h}{\sqrt{2mT_B}} \]
\[ \lambda_A = \frac{h}{\sqrt{2mT_A}} \]

Substituting \( \lambda_B = 2\lambda_A \):
\[ \frac{h}{\sqrt{2mT_B}} = 2 \cdot \frac{h}{\sqrt{2mT_A}} \]

### Step 4: Simplify the equation.
Canceling \( h \) and \( \sqrt{2m} \) from both sides, we have:
\[ \frac{1}{\sqrt{T_B}} = \frac{2}{\sqrt{T_A}} \]

Squaring both sides gives:
\[ \frac{1}{T_B} = \frac{4}{T_A} \]

Rearranging this gives:
\[ T_A = 4T_B \] (Equation 4)

### Step 5: Substitute Equation 4 into Equation 3.
From Equation 3, we have:
\[ T_B = T_A - 1.50 \]
Substituting \( T_A = 4T_B \):
\[ T_B = 4T_B - 1.50 \]

### Step 6: Solve for \( T_B \).
Rearranging gives:
\[ 4T_B - T_B = 1.50 \]
\[ 3T_B = 1.50 \]
\[ T_B = 0.50 \, \text{eV} \]

### Step 7: Find \( T_A \).
Using Equation 4:
\[ T_A = 4T_B = 4 \times 0.50 = 2.00 \, \text{eV} \]

### Step 8: Calculate the work functions \( \phi_A \) and \( \phi_B \).
Substituting \( T_A \) into Equation 1:
\[ \phi_A = 4.25 - 2.00 = 2.25 \, \text{eV} \]

Substituting \( T_B \) into Equation 2:
\[ \phi_B = 4.20 - 0.50 = 3.70 \, \text{eV} \]

### Final Results:
- \( T_A = 2.00 \, \text{eV} \)
- \( T_B = 0.50 \, \text{eV} \)
- \( \phi_A = 2.25 \, \text{eV} \)
- \( \phi_B = 3.70 \, \text{eV} \)