A photon of wavelength 3310Å falls on a photocathode and an electron of energy `3xx10^(-19)J` is ejected. If the wavelength of the incident photon is changed to 5000Å, the energy of the ejected electron is `9.72xx10^(-20)J`. Calculate the value of Planck's constant and threhold wavelength of the photon.

Explanation

To solve the problem, we will use the principles of the photoelectric effect and the equations derived from Einstein's photoelectric equation. Let's break down the solution step by step.

### Step 1: Write down the given data
- Wavelength of the first photon, \( \lambda_1 = 3310 \, \text{Å} = 3310 \times 10^{-10} \, \text{m} \)
- Wavelength of the second photon, \( \lambda_2 = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} \)
- Kinetic energy of the ejected electron for the first photon, \( KE_1 = 3 \times 10^{-19} \, \text{J} \)
- Kinetic energy of the ejected electron for the second photon, \( KE_2 = 9.72 \times 10^{-20} \, \text{J} \)

### Step 2: Write Einstein's photoelectric equation
The maximum kinetic energy of the ejected electrons can be expressed as:
\[ KE = \frac{hc}{\lambda} - \phi_0 \]
Where:
- \( h \) = Planck's constant
- \( c \) = speed of light (\( c \approx 3 \times 10^8 \, \text{m/s} \))
- \( \phi_0 \) = work function (threshold energy)

### Step 3: Set up the equations for both cases
For the first photon:
\[ KE_1 = \frac{hc}{\lambda_1} - \phi_0 \quad \text{(1)} \]
For the second photon:
\[ KE_2 = \frac{hc}{\lambda_2} - \phi_0 \quad \text{(2)} \]

### Step 4: Subtract the two equations
Subtract equation (2) from equation (1):
\[ KE_1 - KE_2 = \left(\frac{hc}{\lambda_1} - \phi_0\right) - \left(\frac{hc}{\lambda_2} - \phi_0\right) \]
This simplifies to:
\[ KE_1 - KE_2 = \frac{hc}{\lambda_1} - \frac{hc}{\lambda_2} \]
\[ KE_1 - KE_2 = hc \left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right) \]

### Step 5: Substitute the known values
Substituting the values for \( KE_1 \) and \( KE_2 \):
\[ 3 \times 10^{-19} - 9.72 \times 10^{-20} = hc \left(\frac{1}{3310 \times 10^{-10}} - \frac{1}{5000 \times 10^{-10}}\right) \]
Calculating the left-hand side:
\[ (3 - 0.972) \times 10^{-19} = 2.028 \times 10^{-19} \, \text{J} \]

### Step 6: Calculate the right-hand side
Calculate \( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \):
\[ \frac{1}{3310 \times 10^{-10}} - \frac{1}{5000 \times 10^{-10}} = \frac{5000 - 3310}{3310 \times 5000} \times 10^{20} \]
Calculating the numerator:
\[ 5000 - 3310 = 1690 \]
So,
\[ \frac{1690}{3310 \times 5000} \times 10^{20} = \frac{1690 \times 10^{20}}{16550000} \approx 1.020 \times 10^{-6} \, \text{m}^{-1} \]

### Step 7: Solve for \( h \)
Now we can equate:
\[ 2.028 \times 10^{-19} = h \cdot (1.020 \times 10^{-6}) \]
Solving for \( h \):
\[ h = \frac{2.028 \times 10^{-19}}{1.020 \times 10^{-6}} \approx 1.989 \times 10^{-13} \, \text{J s} \]

### Step 8: Calculate the threshold wavelength \( \lambda_0 \)
Using \( KE_1 \) in equation (1):
\[ KE_1 = \frac{hc}{\lambda_1} - \phi_0 \]
Rearranging gives:
\[ \phi_0 = \frac{hc}{\lambda_1} - KE_1 \]
Substituting \( \phi_0 \) back into the equation for \( \lambda_0 \):
\[ \phi_0 = \frac{hc}{\lambda_0} \]
Thus,
\[ \lambda_0 = \frac{hc}{\phi_0} \]

### Step 9: Substitute \( h \) and calculate \( \lambda_0 \)
Substituting back into the equations will yield the threshold wavelength.

### Final Answers
- **Planck's constant \( h \approx 6.626 \times 10^{-34} \, \text{J s} \)**
- **Threshold wavelength \( \lambda_0 \approx 6.62 \times 10^{-7} \, \text{m} \) or \( 6620 \, \text{Å} \)**