When photons are incident on a photosensitive | vedclass

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(A) The maximum kinetic energy $(K_{max})$ of the emitted photoelectrons is given by the formula: $K_{max} = \frac{1}{2} m v_{max}^2$.Given $m = 9 	i

Vedclass · Accepted Answer

$1.8$

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(A) The maximum kinetic energy $(K_{max})$ of the emitted photoelectrons is given by the formula: $K_{max} = \frac{1}{2} m v_{max}^2$.Given $m = 9 	imes 10^{-31} 	ext{ kg}$ and $v_{max} = 8 	imes 10^5 	ext{ m/s}$.$K_{max} = \frac{1}{2} 	imes (9 	imes 10^{-31}) 	imes (8 	imes 10^5)^2$.$K_{max} = 0.5 	imes 9 	imes 10^{-31} 	imes 64 	imes 10^{10} = 288 	imes 10^{-21} 	ext{ J}$.To convert this energy into electron-volts (eV),divide by the charge of an electron $(e = 1.6 	imes 10^{-19} 	ext{ C})$:$K_{max} 	ext{ (in eV)} = \frac{288 	imes 10^{-21}}{1.6 	imes 10^{-19}} = 180 	imes 10^{-2} = 1.8 	ext{ eV}$.The stopping potential $(V_s)$ is related to the maximum kinetic energy by the equation: $K_{max} = e V_s$.Therefore,$V_s = \frac{K_{max}}{e} = 1.8 	ext{ V}$.

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