The kinetic energy of the electron emitted when light of frequency `3.5 xx 10^15 H`z falls on a metal surface having threshold frequency `1.5 xx 10^15 H`z is `(h = 6.6 xx 10^-34 Js)`.

Options

A. `1.32 xx 10^-18 J`
B. `3.3 xx 10^-18 J`
C. `6.6 xx 10^-19 J`
D. `1.98 xx 10^-19 J`

Explanation

To find the kinetic energy of the emitted electron when light of frequency \(3.5 \times 10^{15} \, \text{Hz}\) falls on a metal surface with a threshold frequency of \(1.5 \times 10^{15} \, \text{Hz}\), we can use the photoelectric equation:

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

Where:
- \(KE\) is the kinetic energy of the emitted electron,
- \(h\) is Planck's constant (\(6.6 \times 10^{-34} \, \text{Js}\)),
- \(\nu\) is the frequency of the incident light (\(3.5 \times 10^{15} \, \text{Hz}\)),
- \(\nu_0\) is the threshold frequency (\(1.5 \times 10^{15} \, \text{Hz}\)).

### Step 1: Calculate the energy of the incident light
First, we calculate the energy of the incident light using the formula:

\[ E = h\nu \]

Substituting the values:

\[ E = (6.6 \times 10^{-34} \, \text{Js}) \times (3.5 \times 10^{15} \, \text{Hz}) \]

Calculating this gives:

\[ E = 2.31 \times 10^{-18} \, \text{J} \]

### Step 2: Calculate the energy corresponding to the threshold frequency
Next, we calculate the energy corresponding to the threshold frequency:

\[ E_0 = h\nu_0 \]

Substituting the values:

\[ E_0 = (6.6 \times 10^{-34} \, \text{Js}) \times (1.5 \times 10^{15} \, \text{Hz}) \]

Calculating this gives:

\[ E_0 = 9.9 \times 10^{-19} \, \text{J} \]

### Step 3: Calculate the kinetic energy of the emitted electron
Now, we can find the kinetic energy of the emitted electron:

\[ KE = E - E_0 \]

Substituting the values we calculated:

\[ KE = (2.31 \times 10^{-18} \, \text{J}) - (9.9 \times 10^{-19} \, \text{J}) \]

Calculating this gives:

\[ KE = 1.32 \times 10^{-18} \, \text{J} \]

### Final Answer
The kinetic energy of the emitted electron is:

\[ KE = 1.32 \times 10^{-18} \, \text{J} \]