Light of wavelength 3500Å is incident on two metals A and B whose work functions are 3.2 eV and 1.9 eV respectively. Which metal will emit photoelectrons

Options

A. A
B. B
C. both A and B
D. neither A nor B

Explanation

To determine which metal will emit photoelectrons when light of wavelength 3500 Å is incident on them, we need to compare the energy of the incident light with the work functions of the metals.

### Step-by-Step Solution:

1. **Convert the Wavelength to Energy**:
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
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 in meters.

First, convert the wavelength from Ångstroms to meters:
\[ 3500 \, \text{Å} = 3500 \times 10^{-10} \, \text{m} = 3.5 \times 10^{-7} \, \text{m} \]

Now, substitute the values into the energy formula:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{3.5 \times 10^{-7} \, \text{m}} \]
\[ E \approx 5.68 \times 10^{-19} \, \text{J} \]

Convert this energy from joules to electron volts (1 eV = \(1.6 \times 10^{-19} \, \text{J}\)):
\[ E \approx \frac{5.68 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.55 \, \text{eV} \]

2. **Compare with Work Functions**:
- For metal A, the work function \( \phi_A = 3.2 \, \text{eV} \).
- For metal B, the work function \( \phi_B = 1.9 \, \text{eV} \).

Now, we compare the energy of the incident light with the work functions:
- For metal A:
\[ E = 3.55 \, \text{eV} > \phi_A = 3.2 \, \text{eV} \quad \text{(Photoelectrons will be emitted)} \]
- For metal B:
\[ E = 3.55 \, \text{eV} > \phi_B = 1.9 \, \text{eV} \quad \text{(Photoelectrons will also be emitted)} \]

3. **Conclusion**:
Since the energy of the incident light is greater than the work functions of both metals A and B, both metals will emit photoelectrons.

### Final Answer:
Both metals A and B will emit photoelectrons.