Question:
When a photosensitive surface is irradiated by light of wavelengths '$\lambda_1$' and '$\lambda_2$', kinetic energies of emitted photoelectrons are '$E_1$' and '$E_2$' respectively. The work function of the photosensitive surface is
When a photosensitive surface is irradiated by light of wavelengths '$\lambda_1$' and '$\lambda_2$', kinetic energies of emitted photoelectrons are '$E_1$' and '$E_2$' respectively. The work function of the photosensitive surface is
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Multiply energy by its matching wavelength to isolate the constant $hc$: $\lambda E = hc - \lambda W$. Equating the two states and factoring out $W$ yields the correct algebraic ratio in just two quick lines of scratchpad layout!
Updated On: Jun 3, 2026
- $\frac{\lambda_1 E_1 - \lambda_2 E_2}{\lambda_2 - \lambda_1}$
- $\frac{\lambda_1 E_1 + \lambda_2 E_2}{\lambda_2 - \lambda_1}$
- $\frac{\lambda_1 E_2 - \lambda_2 E_1}{\lambda_2 - \lambda_1}$
- $\frac{\lambda_1 E_2 + \lambda_2 E_1}{\lambda_2 - \lambda_1}$
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The Correct Option is A
Solution and Explanation
According to Einstein's photoelectric equation, the maximum kinetic energy of emitted electrons is:
$$E = \frac{hc}{\lambda} - W \implies E + W = \frac{hc}{\lambda} \implies \lambda(E + W) = hc$$
Writing this product relationship for both wavelength states:
$$\lambda_1(E_1 + W) = hc \quad \text{--- (Equation 1)}$$
$$\lambda_2(E_2 + W) = hc \quad \text{--- (Equation 2)}$$
Equating Equation 1 and Equation 2 since both equal the constant product $hc$:
$$\lambda_1(E_1 + W) = \lambda_2(E_2 + W)$$
$$\lambda_1 E_1 + \lambda_1 W = \lambda_2 E_2 + \lambda_2 W$$
Rearranging the terms to group the work function $W$ on one side:
$$\lambda_1 E_1 - \lambda_2 E_2 = \lambda_2 W - \lambda_1 W$$
$$\lambda_1 E_1 - \lambda_2 E_2 = W(\lambda_2 - \lambda_1) \implies W = \frac{\lambda_1 E_1 - \lambda_2 E_2}{\lambda_2 - \lambda_1}$$
Final Answer:
The work function of the surface is represented by option (A).
Final Answer:
The work function of the surface is represented by option (A).
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