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(A) The given electric field equation is $E = E_0 \sin [1.57 	imes 10^7 (x - ct)]$.Comparing this with the standard wave equation $E = E_0 \sin (kx -

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$1.2$

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(A) The given electric field equation is $E = E_0 \sin [1.57 	imes 10^7 (x - ct)]$.Comparing this with the standard wave equation $E = E_0 \sin (kx - \omega t)$,we get the angular frequency $\omega = c 	imes 1.57 	imes 10^7 	ext{ rad/s}$.Since $\omega = 2\pi f$,the frequency $f = \frac{c 	imes 1.57 	imes 10^7}{2\pi}$.The energy of the incident photon is $E_{ph} = hf = \frac{h 	imes c 	imes 1.57 	imes 10^7}{2\pi}$.Substituting the values $h = 6.63 	imes 10^{-34} 	ext{ J s}$ and $c = 3 	imes 10^8 	ext{ m/s}$:$E_{ph} (	ext{in eV}) = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8 	imes 1.57 	imes 10^7}{2 	imes 3.14 	imes 1.6 	imes 10^{-19}} \approx 3.1 	ext{ eV}$.Using Einstein's photoelectric equation: $eV_s = E_{ph} - \Phi$,where $\Phi = 1.9 	ext{ eV}$ is the work function.$eV_s = 3.1 	ext{ eV} - 1.9 	ext{ eV} = 1.2 	ext{ eV}$.Therefore,the stopping potential $V_s = 1.2 	ext{ V}$.

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