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Photoelectric emission is observed from ...

Photoelectric emission is observed from a surface for frequencies `v_1` and `v_2` of the incident radiation `(v_1 gt v_2)`. If the maximum kinetic energies of the photoelectrons in two cases are in ratio `1 : K` then the threshold frequency `v_0` is given by.

A

`(v_2 - v_1)/(K - 1)`

B

`(Kv_1 - v_2)/(K - 1)`

C

`(K v_2 - v_1)/(K - 1)`

D

`(v_2 - v_1)/(K)`

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The correct Answer is:
To solve the problem, we will use the photoelectric effect equation, which relates the energy of the incident photons to the maximum kinetic energy of the emitted photoelectrons. The equation is given by: \[ KE = h\nu - h\nu_0 \] Where: - \( KE \) is the maximum kinetic energy of the emitted photoelectrons, - \( h \) is Planck's constant, - \( \nu \) is the frequency of the incident radiation, - \( \nu_0 \) is the threshold frequency. ### Step-by-Step Solution: 1. **Identify the Maximum Kinetic Energies**: Let the maximum kinetic energy of the photoelectrons for frequency \( v_1 \) be \( KE_1 \) and for frequency \( v_2 \) be \( KE_2 \). According to the problem, we have: \[ \frac{KE_1}{KE_2} = \frac{1}{K} \] This implies: \[ KE_1 = \frac{1}{K} KE_2 \] 2. **Apply the Photoelectric Effect Equation**: For frequency \( v_1 \): \[ KE_1 = h v_1 - h v_0 \] For frequency \( v_2 \): \[ KE_2 = h v_2 - h v_0 \] 3. **Substituting the Kinetic Energies**: Substitute the expressions for \( KE_1 \) and \( KE_2 \): \[ \frac{h v_1 - h v_0}{h v_2 - h v_0} = \frac{1}{K} \] 4. **Simplifying the Equation**: Cancel \( h \) from both sides: \[ \frac{v_1 - v_0}{v_2 - v_0} = \frac{1}{K} \] 5. **Cross-Multiplying**: Cross-multiply to eliminate the fraction: \[ K(v_1 - v_0) = v_2 - v_0 \] 6. **Rearranging the Equation**: Rearranging gives: \[ K v_1 - K v_0 = v_2 - v_0 \] \[ K v_1 - v_2 = K v_0 - v_0 \] \[ K v_1 - v_2 = (K - 1)v_0 \] 7. **Solving for Threshold Frequency \( v_0 \)**: Finally, we can express \( v_0 \): \[ v_0 = \frac{K v_1 - v_2}{K - 1} \] ### Final Result: The threshold frequency \( v_0 \) is given by: \[ v_0 = \frac{K v_1 - v_2}{K - 1} \]

To solve the problem, we will use the photoelectric effect equation, which relates the energy of the incident photons to the maximum kinetic energy of the emitted photoelectrons. The equation is given by: \[ KE = h\nu - h\nu_0 \] Where: - \( KE \) is the maximum kinetic energy of the emitted photoelectrons, - \( h \) is Planck's constant, - \( \nu \) is the frequency of the incident radiation, ...
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