The electric field associated with a monochro | vedclass

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(C) Given the wave equation: $E = E_0 \sin [k(x - ct)]$,where $k = \frac{2\pi}{\lambda} = 1.57 	imes 10^7 	ext{ m}^{-1}$.Calculating wavelength $\la

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$1.1$

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(C) Given the wave equation: $E = E_0 \sin [k(x - ct)]$,where $k = \frac{2\pi}{\lambda} = 1.57 	imes 10^7 	ext{ m}^{-1}$.Calculating wavelength $\lambda$: $\lambda = \frac{2 	imes 3.14}{1.57 	imes 10^7} = 4 	imes 10^{-7} 	ext{ m}$.Energy of incident photon $E_p = \frac{hc}{\lambda} = \frac{6.64 	imes 10^{-34} 	imes 3 	imes 10^8}{4 	imes 10^{-7}} = 4.98 	imes 10^{-19} 	ext{ J}$.Converting work function $\phi_0$ to Joules: $\phi_0 = 1.9 	ext{ eV} = 1.9 	imes 1.6 	imes 10^{-19} 	ext{ J} = 3.04 	imes 10^{-19} 	ext{ J}$.Using Einstein's photoelectric equation: $E_p = \phi_0 + eV_0$.$eV_0 = 4.98 	imes 10^{-19} - 3.04 	imes 10^{-19} = 1.94 	imes 10^{-19} 	ext{ J}$.$V_0 = \frac{1.94 	imes 10^{-19}}{1.6 	imes 10^{-19}} 	ext{ V} = 1.2125 	ext{ V}$.Rounding to the nearest option,the stopping potential is $1.1 	ext{ V}$.

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