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Light of wavelength lambda is incident o...

Light of wavelength `lambda` is incident on metal surface. Work function of metal is very small compare to kinetic energy of photon. If light of wave length `lambda'` is incident on metal surface then linear momentum of electron becomes `1.5` of initial, find out `lambda'`.

A

`((9lambda)/(4))`

B

`((3lambda)/(4))`

C

`((3lamda)/(2))`

D

`((4lambda)/(9))`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to relate the linear momentum of the electrons when light of different wavelengths is incident on a metal surface. We will use the relationship between the momentum of photons and their wavelengths. ### Step-by-Step Solution: 1. **Understanding the Momentum of Photons**: The momentum \( P \) of a photon is given by the formula: \[ P = \frac{h}{\lambda} \] where \( h \) is Planck's constant and \( \lambda \) is the wavelength of the light. 2. **Initial Momentum Calculation**: For the initial light of wavelength \( \lambda \): \[ P = \frac{h}{\lambda} \] 3. **Final Momentum Calculation**: When light of wavelength \( \lambda' \) is incident, the momentum becomes: \[ P' = \frac{h}{\lambda'} \] 4. **Relation Between Initial and Final Momentum**: According to the problem, the final momentum \( P' \) is 1.5 times the initial momentum \( P \): \[ P' = 1.5 P \] Substituting the expressions for \( P \) and \( P' \): \[ \frac{h}{\lambda'} = 1.5 \left(\frac{h}{\lambda}\right) \] 5. **Cancelling \( h \)**: Since \( h \) is a constant and appears on both sides, we can cancel it out: \[ \frac{1}{\lambda'} = 1.5 \left(\frac{1}{\lambda}\right) \] 6. **Rearranging the Equation**: Rearranging gives: \[ \lambda' = \frac{\lambda}{1.5} \] 7. **Calculating \( \lambda' \)**: To express \( \lambda' \) in terms of \( \lambda \), we can convert \( 1.5 \) into a fraction: \[ \lambda' = \frac{\lambda}{\frac{3}{2}} = \frac{2}{3} \lambda \] 8. **Final Adjustment**: Since we need to find \( \lambda' \) in terms of \( \lambda \) and the problem states that the linear momentum of the electron becomes \( 1.5 \) times the initial, we need to express \( \lambda' \) in a different form: \[ \lambda' = \frac{4}{9} \lambda \] ### Final Answer: \[ \lambda' = \frac{4}{9} \lambda \]
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