Ncert Exercises & Solutions

Question 11.28: A mercury lamp is a convenient source for studying the frequency dependence of photoelectric emission since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury sourcæ were used: $\lambda_1=3560~\mathring{A},\lambda_2=4047~\mathring{A},~\lambda_3=4358~\mathring{A},~\lambda_4=5461~\mathring{A},~\lambda_5=6907~\mathring{A}~$The stopping voltages, respectively, were measured to be:

$V_{01} = 1 .28 V, V_{02} = 0 .95 V, V_{03} = 0 .74 V, V_{04} = 0 .16 V, V_{05} = 0 V$

Determine the value of Planck's constant h, the threshold frequency, and work function for the material. [Note: You will notice that to get h from the data, you will need to know e (which you can take to be 1.6 x 10^-19 C). Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of e (from the oil-drop experiment) confirmed Einstein's photoelectric equation and at the same time gave an independent estimate of the value of h.]

Hint: Use Einstein's photoelectric equation.
Step 1: Find the frequencies for various lines.
$\text{Einstein's photoelectric equation is given as:}\\ \text{eV}_{0}=\text{h}\nu-\phi_{0} \\ \text{V}_{0}=\frac{\text{h}}{\text{e}} \text{v}-\frac{\phi_{0}}{\text{e}} \ldots \ldots \ldots \ldots(1)\\ \text{Where,}\\ \text{V}_{0}= \text{Stopping potential}\\ \text{h = Planck's constant}\\ \text{e = Charge on an electron}\\ \nu\text{ = Frequency of radiation}\\ \phi_{0}= \text{Work function of a material}\\ \text{It can be concluded from equation (1) that potential}~ \text{V}_{0} \text{ is directly proportional to frequency }\nu.\\ \text{Frequency is also given by the relation:}\\ \nu=\frac{\text { Speed of light }(c)}{\text { Wavelenght }(\lambda)}$
$\text{This relation can be used to obtain the frequencies of the various lines of the given wavelengths.}\\ \mathrm{\nu}_{1}=\frac{\mathrm{c}}{\lambda_{1}}=\frac{3 \times 10^{8}}{3650 \times 10^{-10}}=8.219 \times 10^{14} \mathrm{~Hz} \\ \mathrm{\nu}_{2}=\frac{\mathrm{c}}{\lambda_{2}}=\frac{3 \times 10^{8}}{4047 \times 10^{-10}}=7.412 \times 10^{14} \mathrm{~Hz} \\ \mathrm{\nu}_{3}=\frac{\mathrm{c}}{\lambda_{3}}=\frac{3 \times 10^{8}}{4358 \times 10^{-10}}=6.884 \times 10^{14} \mathrm{~Hz} \\ \mathrm{\nu}_{4}=\frac{\mathrm{c}}{\lambda_{4}}=\frac{3 \times 10^{8}}{5461 \times 10^{-10}}=5.493 \times 10^{14} \mathrm{~Hz} \\ \mathrm{\nu}_{5}=\frac{\mathrm{c}}{\lambda_{5}}=\frac{3 \times 10^{8}}{6907 \times 10^{-10}}=4.343 \times 10^{14} \mathrm{~Hz}\\ \text{The given quantities can be listed in tabular form as:}$

Step 2: Draw a graph between stopping potential and frequency.
The following figure shows a graph between $\nu$ and V₀.

It can be observed that the obtained curve is a straight line. It intersects the V-axis at 5 × 10¹⁴ Hz, which is the threshold frequency $\nu_0$ of the material. Point D corresponds to a frequency less than the threshold frequency. Hence, there is no photoelectric emission for the $\lambda_{5}$ line, and therefore, no stopping voltage is required to stop the current.

Step 3: Find the slope of the line from the given graph.
$\text{Slope of the straight line }=\frac{A B}{C B}=\frac{1.28-0.16}{(8.214-5.493) \times 10^{14}}\\ \text{From equation (1), the slope}~ \frac{h}{\text{e}}~ \text{can be written as:}\\ \frac{\mathrm{h}}{\mathrm{e}}=\frac{1.28-0.16}{(8.214-5.493) \times 10^{14}} \\ \therefore \mathrm{h}=\frac{1.12 \times 1.6 \times 10^{-19}}{2.726 \times 10^{14}} \\ =6.573 \times 10^{-34 }\mathrm{Js}$

Step 4: Find the work function of the metal.
$\text{The work function of the metal is given as:} \phi_{0}=\mathrm{hv}_{0} \\ =6.573 \times 10^{-34} \times 5 \times 10^{14} \\ =3.286 \times 10^{-19} \mathrm{~J} \\ =\frac{3.286 \times 10^{-19}}{1.6 \times 1^{-18}} \\ =2.054 \mathrm~{eV}$

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