`U.V`. light of wavelength `800A^(@)&700A^(@)` falls on hydrogen atoms in their ground state `&` liberates electrons with kinetic energy `1.8eV` and `4eV` respectively. Calculate planck's constant.

Explanation

To calculate Planck's constant (H) based on the given information, we will follow these steps:

### Step 1: Understand the Energy Equation
The energy of the photons can be expressed using the equation:
\[ E = K.E + W \]
Where:
- \( E \) is the energy of the photon,
- \( K.E \) is the kinetic energy of the liberated electron,
- \( W \) is the work function of the hydrogen atom.

### Step 2: Calculate the Energy of the Photons
The energy of a photon can also be expressed in terms of its wavelength (\( \lambda \)):
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) is Planck's constant,
- \( c \) is the speed of light (approximately \( 3 \times 10^8 \) m/s),
- \( \lambda \) is the wavelength in meters.

### Step 3: Set Up the Equations
For the two wavelengths given:
1. For \( \lambda_1 = 800 \, \text{Å} = 800 \times 10^{-10} \, \text{m} \):
\[ E_1 = K.E_1 + W \]
\[ E_1 = \frac{hc}{800 \times 10^{-10}} \]
\[ K.E_1 = 1.8 \, \text{eV} \]

2. For \( \lambda_2 = 700 \, \text{Å} = 700 \times 10^{-10} \, \text{m} \):
\[ E_2 = K.E_2 + W \]
\[ E_2 = \frac{hc}{700 \times 10^{-10}} \]
\[ K.E_2 = 4 \, \text{eV} \]

### Step 4: Write the Equations
From the above, we can write:
1. \( \frac{hc}{800 \times 10^{-10}} = 1.8 + W \) (Equation 1)
2. \( \frac{hc}{700 \times 10^{-10}} = 4 + W \) (Equation 2)

### Step 5: Subtract the Equations
Subtract Equation 1 from Equation 2 to eliminate \( W \):
\[ \frac{hc}{700 \times 10^{-10}} - \frac{hc}{800 \times 10^{-10}} = (4 - 1.8) \]
\[ \frac{hc}{700 \times 10^{-10}} - \frac{hc}{800 \times 10^{-10}} = 2.2 \]

### Step 6: Factor Out \( hc \)
Factoring out \( hc \):
\[ hc \left( \frac{1}{700 \times 10^{-10}} - \frac{1}{800 \times 10^{-10}} \right) = 2.2 \]

### Step 7: Simplify the Left Side
Calculate the left side:
\[ \frac{1}{700 \times 10^{-10}} - \frac{1}{800 \times 10^{-10}} = \frac{800 - 700}{700 \times 800} \times 10^{10} = \frac{100}{560000} \times 10^{10} \]
\[ = \frac{1}{5600} \times 10^{10} \]

### Step 8: Substitute Back
Now substitute back into the equation:
\[ hc \left( \frac{1}{5600} \times 10^{10} \right) = 2.2 \]
\[ hc = 2.2 \times 5600 \times 10^{-10} \]

### Step 9: Calculate Planck's Constant
Convert \( 2.2 \) eV to Joules (1 eV = \( 1.6 \times 10^{-19} \) J):
\[ hc = 2.2 \times 5600 \times 1.6 \times 10^{-19} \times 10^{-10} \]
\[ hc = 2.2 \times 5600 \times 1.6 \times 10^{-29} \]
\[ hc = 6.57 \times 10^{-34} \, \text{Joule second} \]

### Final Answer
Thus, the value of Planck's constant \( h \) is approximately:
\[ h \approx 6.57 \times 10^{-34} \, \text{Joule second} \]