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  • 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.

Q: 11.28 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\hspace{1mm}\dot{A},\lambda _2=4047\hspace{1mm}\dot{A}, \lambda _3=4358\hspace{1mm}\dot{A}, \lambda _4=5461\hspace{1mm}\dot{A},\lambda _5=6907\hspace{1mm}\dot{A}.

The stopping voltages, respectively, were measured to be

 V _0_1=1.28\hspace{1mm}V,V _0_2=0.95\hspace{1mm}V,V _0_3=0.74\hspace{1mm}V, V_0_4=0.16\hspace{1mm}V,V _0_5=0\hspace{1mm}V.

Determine the value of Planck’s constant  h, the threshold frequency and work function for the material. 

                       

Answers (1)

best_answer

\\h\nu =\phi _{0}+eV\\ V=(\frac{h}{e})\nu -\phi _{0}\\

where V is stopping potential, h is planks constant, e is electronic charge, \nu is frequency of incident photons and \phi _{0} is work function of metal in electron Volts.

To calculate the planks constant from the above date we plot the stopping potential vs frequency graph

\nu_{1}=\frac{c}{\lambda_{1} }=\frac{3\times 10^{8}}{3650\times 10^{-10}}=8.219\times 10^{14}\ Hz

\nu_{2}=\frac{c}{\lambda_{2} }=\frac{3\times 10^{8}}{4047\times 10^{-10}}=7.412\times 10^{14}\ Hz

\nu_{3}=\frac{c}{\lambda_{3} }=\frac{3\times 10^{8}}{4358\times 10^{-10}}=6.884\times 10^{14}\ Hz

\nu_{4}=\frac{c}{\lambda_{4} }=\frac{3\times 10^{8}}{5461\times 10^{-10}}=5.493\times 10^{14}\ Hz

\nu_{5}=\frac{c}{\lambda_{5} }=\frac{3\times 10^{8}}{6907\times 10^{-10}}=4.343\times 10^{14}\ Hz

V _0_1=1.28\hspace{1mm}V,V _0_2=0.95\hspace{1mm}V,V _0_3=0.74\hspace{1mm}V, V_0_4=0.16\hspace{1mm}V,V _0_5=0\hspace{1mm}V.

The plot we get is

From the above figure, we can see that the curve is almost a straight line.

 

The slope of the above graph will give the Plank's constant divided by the electronic charge. Planks constant calculated from the above chart is 

\\h=\frac{\left ( 1.28-0.16 \right )\times 1.6\times 10^{-19}}{(8.214-5.493)\times 10^{14}}\\ h=6.573\times 10^{-34}\ Js

Planks constant calculated from the above chart is therefore 6.573\times 10^{-34}\ Js

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Sayak

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