Question: Q. 3. If light of wavelength $412.5 \mathrm{~nm}$ is incident on each of the metals given below, which ones will show photoelectric emission and why?
| Metal | Work Function (eV) |
|---|---|
| $\mathrm{Na}$ | 1.92 |
| $\mathrm{K}$ | 2.15 |
| $\mathrm{Ca}$ | 3.20 |
| $\mathrm{Mo}$ | 4.17 |
R [OD 2018]
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Solution:
Ans. Calculating the energy of the incident photon 1 Identifying the metals Reason
The energy of a photon of incident radiation is given by
$$ \begin{align*} E & =\frac{h c}{\lambda} \ E & =\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\left(412.5 \times 10^{-9}\right) \times\left(1.6 \times 10^{-19}\right)} \mathrm{eV} \ & =3.01 \mathrm{eV} \end{align*} $$
Hence, only $\mathrm{Na}$ and $\mathrm{K}$ will show photoelectric emission.
[Note: Award this $1 / 2$ mark even if the student writes the name of only one of these metals]
Reason : The energy of the incident photon is more than the work function of only these two metals. $1 / 2$
[CBSE Marking Scheme 2018]
Detailed Answer :
Given, $\lambda=412.5 \mathrm{~nm}=4125 \AA$
$\therefore$ Energy of light, $E_{\lambda}=\frac{12375}{4125} \mathrm{eV}$
$$ E_{\lambda}=3.0 \mathrm{eV} $$
Now for photoelectric emission, we should have energy of incident light higher than the work function of metal. Since work function of $\mathrm{Na}$ and $\mathrm{K}$ is less than $E_{\lambda}$, so these metals $\mathrm{Na}$ and $\mathrm{K}$ show photo electric emission.
But for $\mathrm{Ca}$ and Mo, work function is more than $E_{\lambda^{\prime}}$, so these metals $\mathrm{Ca}$ and Mo, do not show photoelectric emission.
Answering Tips
- Before comparing the work function energy the incident radiation photon energy should be converted in e.V.
