Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • 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 the frequency dependence of photoelectric emission since it gives several 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 the 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 the electronic charge, \nu is the frequency of incident photons and \phi _{0} is the 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.ph

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

Posted by

Sayak

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Board Exams 2025: Counselling for students, parents from February 1
anam-khan

CBSE asks schools to include APAAR IDs in school ID after NTA says ‘not mandatory’ for NEET
anam-khan

CBSE Board Exams 2025: What the ethics code bans from exam centres

Latest Question