Work function of a metal is 2 eV. The maximum wavelength of photons required to emit electrons from its surface is

Options

A. 6215 `Å`
B. 6500 `Å`
C. 5700 `Å`
D. 5900 `Å`

Answer

To find the maximum wavelength of photons required to emit electrons from the surface of a metal with a work function of 2 eV, we can follow these steps:

### Step 1: Understand the Work Function
The work function (φ₀) of a metal is the minimum energy required to remove an electron from its surface. In this case, the work function is given as 2 eV.

### Step 2: Convert Work Function to Joules
Since we need to use SI units, we must convert the work function from electron volts to joules. The conversion factor is:
1 eV = 1.6 × 10⁻¹⁹ J

So, we calculate:
\[ \phi_0 = 2 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 3.2 \times 10^{-19} \, \text{J} \]

### Step 3: Use the Energy-Wavelength Relationship
The energy of a photon (E) is related to its wavelength (λ) by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J s}\))
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\))

### Step 4: Set Up the Equation for Maximum Wavelength
To find the maximum wavelength (λ_max), we set the energy of the photon equal to the work function:
\[ \phi_0 = \frac{hc}{\lambda_{\text{max}}} \]

Rearranging this gives:
\[ \lambda_{\text{max}} = \frac{hc}{\phi_0} \]

### Step 5: Substitute the Values
Now we substitute the known values into the equation:
\[ \lambda_{\text{max}} = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{3.2 \times 10^{-19} \, \text{J}} \]

### Step 6: Calculate λ_max
Calculating the numerator:
\[ hc = 6.626 \times 10^{-34} \times 3 \times 10^8 = 1.9878 \times 10^{-25} \, \text{J m} \]

Now, substituting this into the equation for λ_max:
\[ \lambda_{\text{max}} = \frac{1.9878 \times 10^{-25}}{3.2 \times 10^{-19}} = 6.21125 \times 10^{-7} \, \text{m} \]

### Step 7: Convert to Angstroms
To convert meters to angstroms (1 angstrom = \(10^{-10}\) m):
\[ \lambda_{\text{max}} = 6.21125 \times 10^{-7} \, \text{m} \times 10^{10} \, \text{Å/m} = 6211.25 \, \text{Å} \]

### Step 8: Round to the Nearest Option
The closest answer to our calculation is approximately 6215 Å.

### Final Answer
The maximum wavelength of photons required to emit electrons from the surface of the metal is approximately **6215 Å**.

---