The wavelength of a photon is 1.4Å. It collides with an electron. Its wavelength after collision is 4Å. Calculate the energy of scattered electron. `h=6.62xx10^(-34)Js`

Explanation

To solve the problem step by step, we need to calculate the energy of the scattered electron after the collision with the photon.

### Step 1: Understand the relationship between energy and wavelength
The energy \( E \) of a photon is given by the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (\( 6.62 \times 10^{-34} \, \text{Js} \))
- \( c \) is the speed of light (\( 3 \times 10^{8} \, \text{m/s} \))
- \( \lambda \) is the wavelength of the photon

### Step 2: Calculate the initial energy of the photon
Given the initial wavelength \( \lambda_1 = 1.4 \, \text{Å} = 1.4 \times 10^{-10} \, \text{m} \):
\[ E_1 = \frac{hc}{\lambda_1} = \frac{(6.62 \times 10^{-34} \, \text{Js})(3 \times 10^{8} \, \text{m/s})}{1.4 \times 10^{-10} \, \text{m}} \]

Calculating this gives:
\[ E_1 = \frac{(6.62 \times 3) \times 10^{-26}}{1.4} = \frac{19.86 \times 10^{-26}}{1.4} \approx 1.419 \times 10^{-25} \, \text{J} \]

### Step 3: Calculate the final energy of the photon
Given the final wavelength \( \lambda_2 = 4 \, \text{Å} = 4 \times 10^{-10} \, \text{m} \):
\[ E_2 = \frac{hc}{\lambda_2} = \frac{(6.62 \times 10^{-34} \, \text{Js})(3 \times 10^{8} \, \text{m/s})}{4 \times 10^{-10} \, \text{m}} \]

Calculating this gives:
\[ E_2 = \frac{(6.62 \times 3) \times 10^{-26}}{4} = \frac{19.86 \times 10^{-26}}{4} \approx 4.965 \times 10^{-26} \, \text{J} \]

### Step 4: Calculate the change in energy of the photon
The change in energy \( \Delta E \) is given by:
\[ \Delta E = E_1 - E_2 \]
Substituting the values we calculated:
\[ \Delta E = (1.419 \times 10^{-25} \, \text{J}) - (4.965 \times 10^{-26} \, \text{J}) \approx 9.23 \times 10^{-26} \, \text{J} \]

### Step 5: Conclusion
The energy of the scattered electron is equal to the change in energy of the photon:
\[ \text{Energy of scattered electron} = \Delta E \approx 9.23 \times 10^{-26} \, \text{J} \]

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