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Question

(A) The energy of the incident photon is given by $E = \frac{12400}{\lambda (	ext{\AA})} 	ext{ eV} = \frac{12400}{2000} 	ext{ eV} = 6.2 	ext{ eV}$

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$8.4 	imes 10^5 	ext{ m/s}$

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(A) The energy of the incident photon is given by $E = \frac{12400}{\lambda (	ext{\AA})} 	ext{ eV} = \frac{12400}{2000} 	ext{ eV} = 6.2 	ext{ eV}$.Using Einstein's photoelectric equation: $E = W_0 + K_{\max}$, where $W_0 = 4.2 	ext{ eV}$ is the work function.$K_{\max} = E - W_0 = 6.2 	ext{ eV} - 4.2 	ext{ eV} = 2.0 	ext{ eV}$.Converting to Joules: $K_{\max} = 2.0 	imes 1.6 	imes 10^{-19} 	ext{ J} = 3.2 	imes 10^{-19} 	ext{ J}$.Since $K_{\max} = \frac{1}{2}mv_{\max}^2$, we have $v_{\max} = \sqrt{\frac{2K_{\max}}{m}}$.Substituting $m = 9.1 	imes 10^{-31} 	ext{ kg}$:$v_{\max} = \sqrt{\frac{2 	imes 3.2 	imes 10^{-19}}{9.1 	imes 10^{-31}}} = \sqrt{0.703 	imes 10^{12}} \approx 0.838 	imes 10^6 	ext{ m/s} \approx 8.4 	imes 10^5 	ext{ m/s}$.

Energy required to remove an electron from an aluminium surface is $4.2 \text{ eV}$. If light of wavelength $2000 \text{ \AA}$ falls on the surface, the velocity of the fastest electron ejected from the surface will be:

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