The work function of two metals metal A and metal B are 6.5 eV and 4.5 eV respectively. If the threshold wavelength of metal A is 2500 A, the threshold wavelength on metal B will be approximately equal to

Options

A. 3611A
B. 3500 A
C. 6332 A
D. 4321 A

Answer

To find the threshold wavelength of metal B, we can use the relationship between the work function (φ) and the threshold wavelength (λ) in the context of the photoelectric effect. The work function is the minimum energy required to eject an electron from the surface of a metal, and it is related to the threshold wavelength by the equation:

\[ \phi = \frac{hc}{\lambda} \]

Where:
- \( \phi \) is the work function in electron volts (eV),
- \( h \) 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 threshold wavelength in meters.

Given:
- Work function of metal A, \( \phi_A = 6.5 \, \text{eV} \)
- Work function of metal B, \( \phi_B = 4.5 \, \text{eV} \)
- Threshold wavelength of metal A, \( \lambda_A = 2500 \, \text{Å} = 2500 \times 10^{-10} \, \text{m} \)

### Step 1: Convert Work Functions to Joules
First, we need to convert the work functions from electron volts to joules since \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\).

\[ \phi_A = 6.5 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.04 \times 10^{-18} \, \text{J} \]

\[ \phi_B = 4.5 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 7.2 \times 10^{-19} \, \text{J} \]

### Step 2: Use the Relationship Between Work Function and Wavelength
Since the work function is inversely proportional to the threshold wavelength, we can write:

\[ \frac{\phi_A}{\phi_B} = \frac{\lambda_B}{\lambda_A} \]

### Step 3: Substitute Known Values
Substituting the known values into the equation:

\[ \frac{6.5}{4.5} = \frac{\lambda_B}{2500 \, \text{Å}} \]

### Step 4: Solve for \( \lambda_B \)
Cross-multiplying gives:

\[ \lambda_B = 2500 \, \text{Å} \times \frac{6.5}{4.5} \]

Calculating the fraction:

\[ \frac{6.5}{4.5} = \frac{65}{45} = \frac{13}{9} \]

Now substituting this back:

\[ \lambda_B = 2500 \, \text{Å} \times \frac{13}{9} \]

Calculating \( \lambda_B \):

\[ \lambda_B = 2500 \times 1.4444 \approx 3611 \, \text{Å} \]

### Final Answer
The threshold wavelength of metal B is approximately \( 3611 \, \text{Å} \).

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