CBSE 2025, Compartment, Set 1

Question

The work function of Caesium is $2.14	ext{ eV}$. Find the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of $0.86	ext{ V}$.

Edvaya · Accepted Answer

The correct answer is Subjective. **SOLUTION** Given: * Work function, $\Phi_0 = 2.14	ext{ eV}$ * Stopping potential, $V_0 = 0.86	ext{ V}$ The maximum kinetic energy ($K_{max}$) is related to the stopping potential by: $$K_{max} = eV_0 = 0.86	ext{ eV}$$ Using Einstein's Photoelectric Equation: $$E = K_{max} + \Phi_0$$ $$E = 0.86	ext{ eV} + 2.14	ext{ eV} = 3.00	ext{ eV}$$ To find the wavelength ($\lambda$), we use $E = \frac{hc}{\lambda}$. Converting $E$ to Joules: $$\lambda = \frac{hc}{E}$$ $$\lambda = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{3.00 	imes 1.6 	imes 10^{-19}}$$ $$\lambda = \frac{19.89 	imes 10^{-26}}{4.8 	imes 10^{-19}}$$ $$\lambda \approx 4.14 	imes 10^{-7}	ext{ m}$$ (or $414	ext{ nm}$)

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Physics > Modern Physics > Dual Nature Of Radiation And Matter

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