The threshold frequency for a certain metal for photoelectric effect is `1.7x10^(15)Hz`. When light of frequency `2.2xx10^(15)Hz` is incident on the metal surface, the kinetic energy of the emitted photoelectrons is `3.3xx10^(-19)J`. What is the value of the Planck's constant ?

Options

A. `3.3xx10^(-34)` J-s
B. `6.6xx10^(-34)` J-s
C. `8.6xx10^(-34)` J-s
D. `4xx10^(-35)` J-s

Explanation

To find the value of Planck's constant \( h \), we will use the photoelectric effect equation derived from Einstein's theory:

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

Where:
- \( KE \) is the kinetic energy of the emitted photoelectrons.
- \( \nu \) is the frequency of the incident light.
- \( \nu_0 \) is the threshold frequency of the metal.

### Step 1: Identify the given values
- Threshold frequency \( \nu_0 = 1.7 \times 10^{15} \, \text{Hz} \)
- Frequency of incident light \( \nu = 2.2 \times 10^{15} \, \text{Hz} \)
- Kinetic energy \( KE = 3.3 \times 10^{-19} \, \text{J} \)

### Step 2: Calculate the difference in frequency
Calculate \( \nu - \nu_0 \):

\[ \nu - \nu_0 = 2.2 \times 10^{15} \, \text{Hz} - 1.7 \times 10^{15} \, \text{Hz} = 0.5 \times 10^{15} \, \text{Hz} = 5.0 \times 10^{14} \, \text{Hz} \]

### Step 3: Substitute into the photoelectric equation
Rearranging the photoelectric equation gives us:

\[ h = \frac{KE + h \nu_0}{\nu} \]

Since we need to find \( h \), we can express it as:

\[ h = \frac{KE}{\nu - \nu_0} \]

### Step 4: Substitute the known values
Now substitute the values of \( KE \) and \( \nu - \nu_0 \):

\[ h = \frac{3.3 \times 10^{-19} \, \text{J}}{5.0 \times 10^{14} \, \text{Hz}} \]

### Step 5: Calculate \( h \)
Perform the calculation:

\[ h = \frac{3.3 \times 10^{-19}}{5.0 \times 10^{14}} = 6.6 \times 10^{-34} \, \text{J s} \]

### Final Answer
Thus, the value of Planck's constant \( h \) is:

\[ h \approx 6.6 \times 10^{-34} \, \text{J s} \]