The work function for Ag metal is `7 .5 xx 10^(-19) J`. As metal is being exposed to the light of frequency 1220 `Å` . Which is `//` are correct statements?

Options

A. Threshold frequency of metal is `1/135 xx10^(15)s^(-1)`
B. Threshold frequency of metal is `8. 33 xx 10^(15) s^(-1)`
C. Stopping potential is 5.46 volt
D. If light of wavelength 3600 `Å` is used then photo -electric effect take place

Answer

To solve the problem, we need to determine whether the light of frequency corresponding to a wavelength of 1220 Å (angstrom) can cause the photoelectric effect in silver (Ag) metal, given its work function. We will follow these steps:

### Step 1: Convert Wavelength to Meters
The wavelength is given in angstroms, and we need to convert it to meters for our calculations.

\[ \text{Wavelength} (\lambda) = 1220 \, \text{Å} = 1220 \times 10^{-10} \, \text{m} \]

### Step 2: Calculate the Energy of the Incident Light
We use the formula for the energy of a photon:

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

Where:
- \( h = 6.626 \times 10^{-34} \, \text{J s} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)

Substituting the values:

\[ E = \frac{(6.626 \times 10^{-34} \, \text{J s}) \times (3 \times 10^8 \, \text{m/s})}{1220 \times 10^{-10} \, \text{m}} \]

Calculating the energy:

\[ E = \frac{1.9878 \times 10^{-25}}{1.22 \times 10^{-7}} \approx 1.63 \times 10^{-18} \, \text{J} \]

### Step 3: Compare Energy with Work Function
The work function (\( \phi \)) for silver is given as:

\[ \phi = 7.5 \times 10^{-19} \, \text{J} \]

Now we compare the energy of the incident light with the work function:

\[ E = 1.63 \times 10^{-18} \, \text{J} > \phi = 7.5 \times 10^{-19} \, \text{J} \]

Since the energy of the incident light is greater than the work function, the photoelectric effect will occur.

### Step 4: Calculate Stopping Potential
The stopping potential (\( V_0 \)) can be calculated using the formula:

\[ E - \phi = eV_0 \]

Where \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \, \text{C} \)).

Substituting the values:

\[ 1.63 \times 10^{-18} \, \text{J} - 7.5 \times 10^{-19} \, \text{J} = (1.6 \times 10^{-19} \, \text{C}) V_0 \]

Calculating the left side:

\[ 0.88 \times 10^{-18} \, \text{J} = (1.6 \times 10^{-19} \, \text{C}) V_0 \]

Now, solving for \( V_0 \):

\[ V_0 = \frac{0.88 \times 10^{-18}}{1.6 \times 10^{-19}} \approx 5.5 \, \text{V} \]

### Conclusion
1. The photoelectric effect occurs because the energy of the incident light is greater than the work function.
2. The stopping potential is approximately 5.5 V.