When a yellow light of wavelength `6000Å` from a sodium lamp falls on a certain photocell, a negative potential of `0.30` volt is needed to stop all the electtons from reaching the collector. Then choose correct option (s)

Options

A. Work function of the given photocel is `25.5eV` approximately
B. Work function of the given photocell is `1.8eV` approximately
C. A negative potential of `0.6` Volt will be needed to stop all the electrons if light if of wavelength `lamda=4000Å` is used
D. A negative potential of `1.3` Volt will be needed to stop all the electrons if light of wavelength `lamda=4000Å` is used

Explanation

To solve the problem, we need to find the work function (φ) of the photocell using the photoelectric effect equation. The equation states:

\[ E = \phi + KE \]

Where:
- \( E \) is the energy of the incident photons,
- \( \phi \) is the work function,
- \( KE \) is the kinetic energy of the emitted electrons.

The energy of the photons can be calculated using the formula:

\[ E = \frac{hc}{\lambda} \]

Where:
- \( h \) 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 of the light in meters.

Given:
- Wavelength \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \)
- Stopping potential \( V_0 = 0.30 \, \text{V} \)

1. **Calculate the energy of the incident photons (E):**
\[ E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{6000 \times 10^{-10} \, \text{m}} \]
\[ E = \frac{1.9878 \times 10^{-25}}{6000 \times 10^{-10}} = \frac{1.9878 \times 10^{-25}}{6 \times 10^{-7}} = 3.313 \times 10^{-19} \, \text{J} \]

2. **Convert energy from Joules to electronvolts (eV):**
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
\[ E = \frac{3.313 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 2.07 \, \text{eV} \]

3. **Use the photoelectric equation to find the work function (φ):**
\[ E = \phi + eV_0 \]
\[ 2.07 \, \text{eV} = \phi + (0.30 \, \text{V}) \]
\[ \phi = 2.07 \, \text{eV} - 0.30 \, \text{eV} = 1.77 \, \text{eV} \]

4. **Determine the stopping potential for a different wavelength (4000 Å):**
\[ E' = \frac{hc}{\lambda'} = \frac{12400 \, \text{eV} \cdot \text{Å}}{4000 \, \text{Å}} = 3.1 \, \text{eV} \]
\[ 3.1 \, \text{eV} = \phi + eV_0' \]
\[ V_0' = 3.1 \, \text{eV} - 1.77 \, \text{eV} = 1.33 \, \text{V} \]

Thus, the stopping potential needed for the 4000 Å light is approximately 1.3 V.

### Final Answer:
The work function of the photocell is approximately 1.77 eV, and the stopping potential for 4000 Å light is 1.3 V.