If a metal is irradiated with light of frequency `3 xx 10^(19) sec^(-1)`, electron is emitted with kinetic energy of `6.625 xx 10^(-15)J`. The threshold frequency of the metal is

Options

A. `2 xx 10^(19) sec^(-1)`
B. `1.25 xx 10^(19) sec^(-1)`
C. `6.625 xx 10^(35) sec^(-1)`
D. `6.625 xx 10^(19) sec^(-1)`

Answer

To find the threshold frequency of the metal, we can use the photoelectric effect equation:

\[ E = h \nu = h \nu_0 + KE \]

Where:
- \( E \) is the energy of the incident photon,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)),
- \( \nu \) is the frequency of the incident light,
- \( \nu_0 \) is the threshold frequency,
- \( KE \) is the kinetic energy of the emitted electron.

### Step 1: Calculate the energy of the incident photon
The energy \( E \) of the photon can be calculated using the frequency \( \nu \):

\[ E = h \nu \]

Given:
- \( \nu = 3 \times 10^{19} \, \text{s}^{-1} \)

Substituting the values:

\[ E = (6.626 \times 10^{-34} \, \text{J s}) \times (3 \times 10^{19} \, \text{s}^{-1}) \]

Calculating this gives:

\[ E = 1.9878 \times 10^{-14} \, \text{J} \]

### Step 2: Use the photoelectric equation to find the threshold frequency
Now, we can rearrange the photoelectric effect equation to solve for the threshold frequency \( \nu_0 \):

\[ h \nu_0 = h \nu - KE \]

Substituting the known values:

\[ h \nu_0 = (1.9878 \times 10^{-14} \, \text{J}) - (6.625 \times 10^{-15} \, \text{J}) \]

Calculating the right side:

\[ h \nu_0 = 1.9878 \times 10^{-14} - 6.625 \times 10^{-15} = 1.3253 \times 10^{-14} \, \text{J} \]

### Step 3: Solve for the threshold frequency \( \nu_0 \)
Now, we can find \( \nu_0 \):

\[ \nu_0 = \frac{h \nu_0}{h} = \frac{1.3253 \times 10^{-14} \, \text{J}}{6.626 \times 10^{-34} \, \text{J s}} \]

Calculating this gives:

\[ \nu_0 \approx 2.00 \times 10^{19} \, \text{s}^{-1} \]

### Final Answer
The threshold frequency of the metal is:

\[ \nu_0 \approx 2.00 \times 10^{19} \, \text{s}^{-1} \]