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Question
The threshold frequency for a certain metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission.
Solution
Threshold frequency of the metal, v0 = 3.3 × 1014 Hz
Frequency of light incident on the metal, v = 8.2 × 1014 Hz
Charge on an electron, e = 1.6 × 10−19 C
Planck’s constant, h = 6.626 × 10−34 Js
Cut-off voltage for the photoelectric emission from the metal = V0
The equation for the cut-off energy is given as:
`"eV"_0 = "h"("v" - "v"_0)`
`"V"_0 = ("h"("v" - "v"_0))/"e"`
= `(6.626 xx 10^(-34) xx (8.2 xx 10^14 - 3.3 xx 10^(14)))/(1.6 xx 10^(-19))`
= 2.0292 V
Therefore, the cut-off voltage for the photoelectric emission is 2.0292 V.
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