The work functions of metals are given as Na | vedclass

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(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$. Using $hc \approx 12400 \, eV \cdot \mathring{A}$,we have $E = \frac{12400

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$Mo$ and $Ni$

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(A) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$. Using $hc \approx 12400 \, eV \cdot \mathring{A}$,we have $E = \frac{12400}{3300} \approx 3.76 \, eV$. Photoelectric emission occurs if the incident photon energy $E$ is greater than the work function $\phi$ of the metal $(E > \phi)$. Alternatively,the threshold wavelength is $\lambda_0 = \frac{hc}{\phi}$. Emission occurs if $\lambda Calculating $\lambda_0$ for each metal: For $Na$: $\lambda_0 = \frac{12400}{1.92} \approx 6458 \, \mathring{A}$. Since $3300 For $K$: $\lambda_0 = \frac{12400}{2.15} \approx 5767 \, \mathring{A}$. Since $3300 For $Mo$: $\lambda_0 = \frac{12400}{4.17} \approx 2973 \, \mathring{A}$. Since $3300 > 2973$,no emission occurs. For $Ni$: $\lambda_0 = \frac{12400}{5.0} = 2480 \, \mathring{A}$. Since $3300 > 2480$,no emission occurs. Thus,$Mo$ and $Ni$ will not show photoelectric emission.

The work functions of metals are given as $Na = 1.92 \, eV$,$K = 2.15 \, eV$,$Mo = 4.17 \, eV$,and $Ni = 5.0 \, eV$. Which of these metals will not show photoelectric emission for radiation of wavelength $3300 \, \mathring{A}$ from a $He-Cd$ laser?

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