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The photoelectric current in a photoelectric cell can be reduced to zero by a stopping potential of 1.8 volt. Monochromatic light of wavelength 2200Å is incident on the cathode. Find the maximum kinetic energy of the photoelectrons in joules. [Charge on electron = 1.6 x 10^{-19} C]

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#### Solution

Given:- V_{0} = 1.8 V, e = 1.6 x 10^{-19} C, λ = 2200 Å

To find:- Maximim kinetic energy (K.E.)_{max}

Formula:- (K.E.)_{max} = eV_{0}

Calculation: Using formula,

(K.E.)_{max} = 1.6 x 10^{-19} x 1.8

∴ (K.E)max = 2.88 x 10^{-19} J

**Maximum kinetic energy of emitted photoelectron is 2.88 x 10 ^{-19} J.**

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