White light consisting of wavelengths `380 nm le lambda le 750` nm is incident on a lead surface. For which one of the following ranges of wavelength will photoelectrons be emitted from the elad surface that has a work funcation `W_(0) = 6.63 xx 10^(-29) J`?

Options

A. `380` nm `le lambda le 750` nm
B. `380` nm `le lambda le 630` nm
C. `380` nm `le lambda le 540` nm
D. No photoelectrons will be emitted

Explanation

To solve the problem, we need to determine the range of wavelengths for which photoelectrons will be emitted from a lead surface with a given work function. The work function \( W_0 \) is given as \( 6.63 \times 10^{-29} \) J.

### Step-by-Step Solution:

1. **Understand the Photoelectric Effect**:
The photoelectric effect states that for photoelectrons to be emitted from a surface, the energy of the incident photons must be greater than or equal to the work function of the material.

2. **Relate Energy to Wavelength**:
The energy \( E \) of a photon can be expressed in terms of its wavelength \( \lambda \) using the formula:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \) J·s) and \( c \) is the speed of light (\( 3 \times 10^8 \) m/s).

3. **Set Up the Inequality**:
For photoelectrons to be emitted, the energy of the photon must be greater than the work function:
\[ \frac{hc}{\lambda} \geq W_0 \]

4. **Rearranging the Equation**:
Rearranging the inequality gives us:
\[ \lambda \leq \frac{hc}{W_0} \]

5. **Calculate the Maximum Wavelength**:
Substitute the values of \( h \), \( c \), and \( W_0 \):
\[ \lambda_{\text{max}} = \frac{(6.63 \times 10^{-34} \, \text{J·s})(3 \times 10^8 \, \text{m/s})}{6.63 \times 10^{-29} \, \text{J}} \]
Simplifying this gives:
\[ \lambda_{\text{max}} = \frac{1.989 \times 10^{-25}}{6.63 \times 10^{-29}} \approx 300 \, \text{nm} \]

6. **Determine the Range of Wavelengths**:
Since the maximum wavelength for which photoelectrons can be emitted is \( 300 \, \text{nm} \), we conclude that any wavelength greater than \( 300 \, \text{nm} \) will not result in photoelectron emission.

7. **Check the Given Wavelength Range**:
The problem states that white light with wavelengths from \( 380 \, \text{nm} \) to \( 750 \, \text{nm} \) is incident on the lead surface. Since \( 380 \, \text{nm} \) is greater than \( 300 \, \text{nm} \), no photoelectrons will be emitted.

8. **Final Conclusion**:
Therefore, the correct answer is that no photoelectrons will be emitted from the lead surface when white light is incident on it.