Radiation of wavelength 546 nm falls on a photo cathode and electrons with maximum kinetic energy of 0.18 eV are emitted. When radiation of wavelength 185 nm falls on the same surface, a (negative) stopping potential of 4.6 V has to be applied to the collector cathode to reduce the photoelectric current to zero. Then, the ratio `(h)/(e)` is

Options

A. `6.6xx10^(-15)JsC^(-1)`
B. `4.14xx10^(-15)JsC^(-1)`
C. `6.6xx10^(-34)JsC^(-1)`
D. `4.12xx10^(-34)JsC^(-1)`

Explanation

To solve the problem, we will use the principles of the photoelectric effect, specifically Einstein's photoelectric equation. The equation states:

\[ E = W + KE_{max} \]

Where:
- \( E \) is the energy of the incident photons,
- \( W \) is the work function of the material,
- \( KE_{max} \) is the maximum kinetic energy of the emitted electrons.

### Step 1: Calculate the energy of the incident radiation for both wavelengths.

The energy of a photon can be calculated using the formula:

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

Where:
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the radiation.

For the first wavelength \( \lambda_1 = 546 \, \text{nm} = 546 \times 10^{-9} \, \text{m} \):

\[ E_1 = \frac{hc}{\lambda_1} \]

For the second wavelength \( \lambda_2 = 185 \, \text{nm} = 185 \times 10^{-9} \, \text{m} \):

\[ E_2 = \frac{hc}{\lambda_2} \]

### Step 2: Set up the equations using the photoelectric effect.

For the first wavelength (546 nm), the equation becomes:

\[ E_1 = W + KE_{max1} \]
\[ \frac{hc}{546 \times 10^{-9}} = W + 0.18 \, \text{eV} \]

For the second wavelength (185 nm), the equation becomes:

\[ E_2 = W + eV_s \]
\[ \frac{hc}{185 \times 10^{-9}} = W + 4.6 \, \text{eV} \]

### Step 3: Subtract the two equations.

Subtract the first equation from the second:

\[ \frac{hc}{185 \times 10^{-9}} - \frac{hc}{546 \times 10^{-9}} = (W + 4.6) - (W + 0.18) \]

This simplifies to:

\[ \frac{hc}{185 \times 10^{-9}} - \frac{hc}{546 \times 10^{-9}} = 4.6 - 0.18 \]

### Step 4: Factor out \( hc \).

Factoring \( hc \) out from the left side gives:

\[ hc \left( \frac{1}{185 \times 10^{-9}} - \frac{1}{546 \times 10^{-9}} \right) = 4.42 \, \text{eV} \]

### Step 5: Solve for \( \frac{h}{e} \).

Rearranging the equation to find \( \frac{h}{e} \):

\[ \frac{h}{e} = \frac{4.42 \, \text{eV}}{\left( \frac{1}{185 \times 10^{-9}} - \frac{1}{546 \times 10^{-9}} \right)} \]

### Step 6: Calculate the values.

Now, substituting the values:

1. Calculate \( \frac{1}{185 \times 10^{-9}} \) and \( \frac{1}{546 \times 10^{-9}} \).
2. Find the difference.
3. Substitute back to find \( \frac{h}{e} \).

### Final Calculation:

After performing the calculations, we find:

\[ \frac{h}{e} \approx 4.12 \times 10^{-15} \, \text{Joule second per Coulomb} \]

### Conclusion:

The ratio \( \frac{h}{e} \) is approximately \( 4.12 \times 10^{-15} \, \text{Joule second per Coulomb} \).