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Question

(B) The threshold wavelength $\lambda_0$ is $8000 \ \mathring{A} = 8000 	imes 10^{-10} \ m$. The work function $\Phi = \frac{hc}{\lambda_0} = \frac{6

Vedclass · Accepted Answer

$5.74 	imes 10^5 \ ms^{-1}$

Vedclass · Answer

(B) The threshold wavelength $\lambda_0$ is $8000 \ \mathring{A} = 8000 	imes 10^{-10} \ m$. The work function $\Phi = \frac{hc}{\lambda_0} = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{8000 	imes 10^{-10}} = 2.475 	imes 10^{-19} \ J$. When exposed to $\lambda = 5000 \ \mathring{A} = 5000 	imes 10^{-10} \ m$,the energy of incident radiation is $E = \frac{hc}{\lambda} = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{5000 	imes 10^{-10}} = 3.96 	imes 10^{-19} \ J$. The kinetic energy $K.E. = E - \Phi = 3.96 	imes 10^{-19} - 2.475 	imes 10^{-19} = 1.485 	imes 10^{-19} \ J$. Using $K.E. = \frac{1}{2} m_e v^2$,we get $v = \sqrt{\frac{2 	imes K.E.}{m_e}} = \sqrt{\frac{2 	imes 1.485 	imes 10^{-19}}{9.1 	imes 10^{-31}}} = \sqrt{0.326 	imes 10^{12}} \approx 5.71 	imes 10^5 \ ms^{-1}$. Rounding to the nearest option,the answer is $5.74 	imes 10^5 \ ms^{-1}$.

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