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(D) Given: Work function $\Phi_0 = 1.6 	imes 10^{-19} 	ext{ J}$,Wavelength $\lambda = 6400 \ \mathring{A} = 6400 	imes 10^{-10} 	ext{ m}$.Accordin

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$1.5 	imes 10^{-19}$

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(D) Given: Work function $\Phi_0 = 1.6 	imes 10^{-19} 	ext{ J}$,Wavelength $\lambda = 6400 \ \mathring{A} = 6400 	imes 10^{-10} 	ext{ m}$.According to Einstein's photoelectric equation: $K_{max} = E - \Phi_0 = \frac{hc}{\lambda} - \Phi_0$.Using $h = 6.63 	imes 10^{-34} 	ext{ J s}$ and $c = 3 	imes 10^8 	ext{ m/s}$:$E = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{6400 	imes 10^{-10}} 	ext{ J} = \frac{19.89 	imes 10^{-26}}{6.4 	imes 10^{-7}} 	ext{ J} \approx 3.107 	imes 10^{-19} 	ext{ J}$.$K_{max} = 3.107 	imes 10^{-19} - 1.6 	imes 10^{-19} 	ext{ J} = 1.507 	imes 10^{-19} 	ext{ J}$.Rounding to the nearest option,$K_{max} \approx 1.5 	imes 10^{-19} 	ext{ J}$.

The work function of a metal surface is $1.6 \times 10^{-19} \text{ J}$. When light of wavelength $6400 \ \mathring{A}$ is incident on this surface,the maximum kinetic energy of the emitted photoelectrons is ....... $\text{J}$.

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