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Question: Answered & Verified by Expert
The work function for the following metals is given: Na: $2.75 \mathrm{eV} ; \mathrm{K}: 2.30 \mathrm{eV} ; \mathrm{Mo}: 4.17 \mathrm{eV}$; Ni: $5.15 \mathrm{eV}$. Which of these metals will not give photoelectric emission for a radiation of wavelength $3300 Å$ from a He-Cd laser placed $1 \mathrm{~m}$ away from the photocell? What happens if the laser is brought nearer and placed $50 \mathrm{~cm}$ away?
PhysicsDual Nature of Matter
Solution:
1036 Upvotes Verified Answer
Given : $\lambda=3300 Å=3300 \times 10^{-10} \mathrm{~m}$
$\therefore \quad$ Energy of photon $\mathrm{E}$
$$
\begin{aligned}
&=\frac{\mathrm{hc}}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{3300 \times 10^{-10}} \\
&=6.02 \times 10^{-19} \mathrm{~J}=\frac{6.02 \times 10^{-19}}{1.6 \times 10^{-19}}=3.76 \mathrm{eV} .
\end{aligned}
$$
Since $\phi_0$ (work function) $>\mathrm{h} v_0$ (incident energy) for Mo and Ni they will not show photo electric effect but photoemission will occur for $\mathrm{Na}$ and $\mathrm{K}$. If laser is brought nearer intensity will increase but not frequency, hence photo current will increase in $\mathrm{Na}$ and $\mathrm{K}$ but there will still be no change for Mo and $\mathrm{Ni}$.

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