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Dual Nature of Radiation and Matter — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsDual Nature of Radiation and Matter5 Marks
Question
A mercury lamp is a convenient source for studying 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 source were used: $\lambda_1=3650\ \mathring{\text{A}},\ \lambda_2=4047\ \mathring{\text{A}},\ \lambda_3=4358\ \mathring{\text{A}},\ \lambda_4=5461\ \mathring{\text{A}},\ \lambda_5=6907\ \mathring{\text{A}},$ 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 $1.6 \times 10^{–19}$ 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.]
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Answer


Einstein's photoelectric equation is given as:
$\text{eV}_\circ=\text{hv}-\phi_\circ$
$\text{V}_\circ=\frac{\text{h}}{\text{e}}\text{v}-\frac{\phi_\circ}{\text{e}}....(1)$
Where,
$\text{V}_\circ$ = Stopping potential
h = Planck's constant
e = Charge on an electron
$\phi_\circ$ = Work function of a material
It can be concluded from equation (1) that potential $\text{V}_\circ$ is directly proportional to frequency v.
Frequency is also given by the relation:
$\text{v}=\frac{\text{Speed of light (c)}}{\text{Wavelength}\ (\lambda)}$
This relation can be used to obtain the frequencies of the various lines of the given wavelengths.
$\text{v}_1=\frac{\text{c}}{\lambda_1}=\frac{3\times10^8}{3650\times10^{-10}}=8.219\times10^{14}\ \text{Hz}$
$\text{v}_2=\frac{\text{c}}{\lambda_2}=\frac{3\times10^8}{4047\times10^{-10}}=7.412\times10^{14}\ \text{Hz}$
$\text{v}_3=\frac{\text{c}}{\lambda_3}=\frac{3\times10^8}{4358\times10^{-10}}=6.884\times10^{14}\ \text{Hz}$
$\text{v}_4=\frac{\text{c}}{\lambda_4}=\frac{3\times10^8}{5461\times10^{-10}}=5.493\times10^{14}\ \text{Hz}$
$\text{v}_5=\frac{\text{c}}{\lambda_5}=\frac{3\times10^8}{6907\times10^{-10}}=4.343\times10^{14}\ \text{Hz}$
The given quantities can be listed in tabular form as:
Frequency $\times 10^{14}$ Hz 8.219 7.412 6.884 5.493 4.343
Stopping potential $\text{V}_\circ$ 1.28 0.95 0.74 0.16 0
The following figure shows a graph between vand $\text{V}_\circ.$

It can be observed that the obtained curve is a straight line. It intersects the v-axis at $5 \times 10^{14}$ Hz, which is the threshold frequency $(\text{V}_\circ)$ 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.
Slope of the straight line $=\frac{\text{AB}}{\text{CB}}=\frac{1.28-0.16}{(8.214-5.493)\times10^{14}}$
From equation (1), the slope $\frac{\text{h}}{\text{e}}$ can be written as:
$\frac{\text{h}}{\text{e}}=\frac{1.28-0.16}{(8.214-5.493)\times10^{14}}$
$\therefore\ \text{h}=\frac{1.12\times1.6\times10^{19}}{2.726\times10^{14}}$
$= 6.573 \times 10^{-34} Js$
The work function of the metal is given as:
$\phi_\circ=\text{hv}_\circ$
$= 6.573 \times 10^{-34} \times 5 \times 10^{14}$
$= 3.286 \times 10^{-19} J$
$=\frac{3.286\times10^{-19}}{1.6\times10^{-18}} $
$= 0.054 eV$

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