The threshold frequency for a certain metal is 3.3 x 10¹⁴ Hz. If light of frequency 8.2 x 10¹⁴ Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.

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Solution

Given,
Threshold frequency, v₀ = 3.3 x 10¹⁴ Hz
Frequency of light, v = 8.2 x 10¹⁴ Hz
Planck's constant, h = 6.63 x 10^–34 Js

Using Einstein's photoelectric equation,

\frac{1}{2} {mv}^{2}_{\max} = h (v - v_{0}) = {eV}_{0}

\Rightarrow

V_{0} = \frac{h (v - v_{0})}{e}

\Rightarrow

V_{0} = \frac{6.62 \times 10^{- 34}}{1.6 \times 10^{- 19}} (8.2 \times 10^{14} - 3.3 \times 10^{14})

i.e.,

V_{0} = 2.03 V .

is the required cut=off voltage for photoelectric emission.

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The threshold frequency for a certain metal is 3.3 x 10¹⁴ Hz. If light of frequency 8.2 x 10¹⁴ Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.