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(D) Given: Work-function, $\phi_{0} = 1 eV = 1.6 	imes 10^{-19} J$.Wavelength, $\lambda = 3000 	ext{Å} = 3000 	imes 10^{-10} m$.The energy of the i

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$1 	imes 10^{6} 	ext{ ms}^{-1}$

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(D) Given: Work-function, $\phi_{0} = 1 eV = 1.6 	imes 10^{-19} J$.Wavelength, $\lambda = 3000 	ext{Å} = 3000 	imes 10^{-10} m$.The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.$E = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^{8}}{3000 	imes 10^{-10}} = 6.63 	imes 10^{-19} J$.According to Einstein's photoelectric equation, $E = \phi_{0} + KE_{max}$.$KE_{max} = E - \phi_{0} = 6.63 	imes 10^{-19} - 1.6 	imes 10^{-19} = 5.03 	imes 10^{-19} J$.Since $KE_{max} = \frac{1}{2}mv^{2}$, we have $v = \sqrt{\frac{2 	imes KE_{max}}{m}}$.$v = \sqrt{\frac{2 	imes 5.03 	imes 10^{-19}}{9.1 	imes 10^{-31}}} = \sqrt{1.105 	imes 10^{12}} \approx 1.05 	imes 10^{6} m/s$.Thus, the velocity is approximately $1 	imes 10^{6} m/s$.

The work-function of a metal is $1 eV$. Light of wavelength $3000 \text{Å}$ is incident on this metal surface. The velocity of emitted photoelectrons will be

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