Photoelectric effect takes place in element A. Its work funcation is `2.5 eV` and threshold wavelength is `lambda`. An other element B is having work function of `5 eV`. Then find out the wavelength that can produce photoelectric effect in B.

Options

A. `lambda//2`
B. `2lambda`
C. `lambda`
D. `3lambda`

Explanation

To solve the problem regarding the photoelectric effect in elements A and B, we will follow these steps:

### Step 1: Understand the Relationship Between Work Function and Wavelength
The work function (φ) of a material is the minimum energy required to eject an electron from its surface. The energy of a photon is related to its wavelength (λ) by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength.

### Step 2: Calculate the Threshold Wavelength for Element A
Given that the work function of element A is \( 2.5 \, \text{eV} \), we can convert this energy into joules (since \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \)):
\[ \phi_A = 2.5 \, \text{eV} = 2.5 \times 1.6 \times 10^{-19} \, \text{J} = 4.0 \times 10^{-19} \, \text{J} \]

Using the energy-wavelength relationship:
\[ \phi_A = \frac{hc}{\lambda_A} \]
Rearranging gives:
\[ \lambda_A = \frac{hc}{\phi_A} \]

### Step 3: Calculate the Threshold Wavelength for Element B
For element B, the work function is \( 5 \, \text{eV} \):
\[ \phi_B = 5 \, \text{eV} = 5 \times 1.6 \times 10^{-19} \, \text{J} = 8.0 \times 10^{-19} \, \text{J} \]

Using the same relationship:
\[ \phi_B = \frac{hc}{\lambda_B} \]
Rearranging gives:
\[ \lambda_B = \frac{hc}{\phi_B} \]

### Step 4: Establish the Ratio of Wavelengths
Since we know that the work function is inversely proportional to the wavelength, we can set up the ratio:
\[ \frac{\phi_A}{\phi_B} = \frac{\lambda_B}{\lambda_A} \]
Substituting the known values:
\[ \frac{2.5 \, \text{eV}}{5 \, \text{eV}} = \frac{\lambda_B}{\lambda} \]

### Step 5: Solve for Wavelength of Element B
From the ratio:
\[ \frac{1}{2} = \frac{\lambda_B}{\lambda} \]
This implies:
\[ \lambda_B = \frac{\lambda}{2} \]

### Conclusion
The wavelength that can produce the photoelectric effect in element B is half of the threshold wavelength of element A.

### Final Answer
\[ \lambda_B = \frac{\lambda}{2} \]