White light consisting of wavelengths `380 nmlelambdale750 nm` is incident on a lead surface. For which one of the following ranges of wavelengths will photoelectrons be emitted from the lead surface that has a work function `W_(0) = 6.63 xx 10^(-19) J`?

Options

A. `380 nmlelambdale750 nm`
B. `380 nmlelambdale410 nm`
C. `380 nmlelambdale630 nm`
D. No photoelectrons will be emitted

Answer

To solve the problem, we need to determine the range of wavelengths of white light that can cause the emission of photoelectrons from a lead surface with a given work function. The work function \( W_0 \) is given as \( 6.63 \times 10^{-19} \) J.

### Step 1: Convert the work function from Joules to electron volts (eV)
The conversion factor between Joules and electron volts is:
\[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \]

To convert the work function:
\[ W_0 = \frac{6.63 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} \approx 4.14 \text{ eV} \]

### Step 2: Calculate the energy of the incident light for the given wavelengths
The energy \( E \) of a photon is given by the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) (Planck's constant) = \( 6.63 \times 10^{-34} \text{ J s} \)
- \( c \) (speed of light) = \( 3 \times 10^8 \text{ m/s} \)
- \( \lambda \) is the wavelength in meters.

### Step 3: Calculate the energy for the minimum wavelength (380 nm)
Convert \( 380 \text{ nm} \) to meters:
\[ \lambda = 380 \text{ nm} = 380 \times 10^{-9} \text{ m} \]
Now, calculate the energy:
\[ E = \frac{(6.63 \times 10^{-34} \text{ J s})(3 \times 10^8 \text{ m/s})}{380 \times 10^{-9} \text{ m}} \approx 5.22 \times 10^{-19} \text{ J} \]
Convert this energy to electron volts:
\[ E \approx \frac{5.22 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} \approx 3.26 \text{ eV} \]

### Step 4: Calculate the energy for the maximum wavelength (750 nm)
Convert \( 750 \text{ nm} \) to meters:
\[ \lambda = 750 \text{ nm} = 750 \times 10^{-9} \text{ m} \]
Now, calculate the energy:
\[ E = \frac{(6.63 \times 10^{-34} \text{ J s})(3 \times 10^8 \text{ m/s})}{750 \times 10^{-9} \text{ m}} \approx 2.65 \times 10^{-19} \text{ J} \]
Convert this energy to electron volts:
\[ E \approx \frac{2.65 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} \approx 1.66 \text{ eV} \]

### Step 5: Compare the energies with the work function
- The energy of the light at \( 380 \text{ nm} \) is approximately \( 3.26 \text{ eV} \).
- The energy of the light at \( 750 \text{ nm} \) is approximately \( 1.66 \text{ eV} \).
- The work function \( W_0 \) is \( 4.14 \text{ eV} \).

Since both energies (3.26 eV and 1.66 eV) are less than the work function (4.14 eV), no photoelectrons will be emitted from the lead surface.

### Conclusion
The answer is that photoelectrons will **not** be emitted for any wavelength in the range of \( 380 \text{ nm} \) to \( 750 \text{ nm} \).