Work function of a metal is `2.1 eV`. Which of the waves of the following wavelengths will be able to emit photoelectrons from its surface ?

Options

A. `4000 Å , 7500 Å`
B. `5500 Å,6000 Å`
C. `4000 Å,6000 Å`
D. None of these

Explanation

To determine which wavelengths of light can emit photoelectrons from a metal with a work function of 2.1 eV, we will follow these steps:

### Step 1: Understand the Work Function
The work function (φ) is the minimum energy required to remove an electron from the surface of a metal. In this case, φ = 2.1 eV.

### Step 2: Convert Work Function to Joules
Since energy can be expressed in joules, we need to convert the work function from electron volts to joules. The conversion factor is:
1 eV = 1.6 × 10^-19 J.

So,
\[ \phi = 2.1 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 3.36 \times 10^{-19} \text{ J}. \]

### Step 3: Use the Energy-Wavelength Relationship
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 (\(6.626 \times 10^{-34} \text{ J s}\)),
- \(c\) is the speed of light (\(3 \times 10^8 \text{ m/s}\)),
- \(\lambda\) is the wavelength in meters.

### Step 4: Set Up the Inequality
To emit photoelectrons, the energy of the photon must be greater than or equal to the work function:
\[ \frac{hc}{\lambda} \geq \phi. \]

### Step 5: Solve for Wavelength
Rearranging the inequality gives:
\[ \lambda \leq \frac{hc}{\phi}. \]

### Step 6: Calculate the Maximum Wavelength
Substituting the values:
\[ \lambda \leq \frac{(6.626 \times 10^{-34} \text{ J s})(3 \times 10^8 \text{ m/s})}{3.36 \times 10^{-19} \text{ J}}. \]
Calculating this gives:
\[ \lambda \leq \frac{1.9878 \times 10^{-25}}{3.36 \times 10^{-19}} \approx 5.91 \times 10^{-7} \text{ m} = 591 \text{ nm}. \]

### Step 7: Compare Given Wavelengths
Now, we need to check the provided wavelengths against the calculated maximum wavelength of 591 nm. Any wavelength shorter than this will be able to emit photoelectrons.

### Step 8: Conclusion
After comparing the given wavelengths, we can determine which wavelengths are less than 591 nm and thus can emit photoelectrons.