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 \(=1.6 \times 10^{-19} C\)]
(Oct. 2015)
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Given:-$ V_0 = 1.8 V, e = 1.6 x 10^{-19} C, \lambda = 2200 Å$
To find:- Maximim kinetic energy $(K.E.)_{max}$
Formula:$- (K.E.)_{max} = eV_0$
Calculation: Using formula,
$(K.E.)_{max} = 1.6 x 10^{-19} x 1.8$
$\therefore (K.E)max = 2.88 x 10^{-19} J$
Maximum kinetic energy of emitted photoelectron is $2.88 x 10^{-19} J.$
To find:- Maximim kinetic energy $(K.E.)_{max}$
Formula:$- (K.E.)_{max} = eV_0$
Calculation: Using formula,
$(K.E.)_{max} = 1.6 x 10^{-19} x 1.8$
$\therefore (K.E)max = 2.88 x 10^{-19} J$
Maximum kinetic energy of emitted photoelectron is $2.88 x 10^{-19} J.$
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