CBSE 2025, Set 1

Question

The threshold frequency for a given metal is $3\cdot6 	imes 10^{14}$ Hz. If monochromatic radiations of frequency $6\cdot8 	imes 10^{14}$ Hz are incident on this metal, find the cut-off potential for the photoelectrons.

Edvaya · Accepted Answer

The correct answer is Subjective. Using Einstein's photoelectric equation: $$eV_0 = h(
u - 
u_0)$$ where: - $h = 6\cdot63 	imes 10^{-34}$ J s - $
u = 6\cdot8 	imes 10^{14}$ Hz (incident frequency) - $
u_0 = 3\cdot6 	imes 10^{14}$ Hz (threshold frequency) - $e = 1\cdot6 	imes 10^{-19}$ C **Calculating:** $$eV_0 = 6\cdot63 	imes 10^{-34} 	imes (6\cdot8 - 3\cdot6) 	imes 10^{14}$$ $$= 6\cdot63 	imes 10^{-34} 	imes 3\cdot2 	imes 10^{14}$$ $$= 21\cdot216 	imes 10^{-20} 	ext{ J}$$ $$V_0 = \frac{21\cdot216 	imes 10^{-20}}{1\cdot6 	imes 10^{-19}} = \frac{21\cdot216}{16} \approx \boxed{1\cdot32 	ext{ V}}$$ > **Key formula:** The cut-off (stopping) potential $V_0$ is the minimum retarding potential needed to stop the most energetic photoelectrons. It depends only on frequency, NOT on intensity. The work function $\phi_0 = h
u_0 = 6\cdot63 	imes 10^{-34} 	imes 3\cdot6 	imes 10^{14} \approx 2\cdot39 	imes 10^{-19}$ J $\approx 1\cdot49$ eV.

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Physics > Modern Physics > Dual Nature Of Radiation And Matter

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