InterviewSolution
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InterviewSolution
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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 and end of the visible spectrum. In our experiment with rubidium photocell, the following lines from a mercury source were used: lamda_(1)=3650Å, lamda_(2)=4047Å,lamda_(3)=4358Å,lamda_(4)=5461Å,lamda_(5)=6907Å The stopping voltages, respectively, were measured to be: V_(01)=1.28V,V_(02)=0.95V,V_(03)=0.74V,V_(04)=0.16V,V_(05)=0V Determine the value of Planck's constant h, the threshold frequency and work function for the material. |
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Answer» Solution :Frequencies corresponding to different wavelengths of spectral lines of mercury are `v_(1)=(c)/(lamda_(1))=(3xx10^(8))/(3650xx10^(-10))=8.219xx10^(14)Hz`  `v_(2)=(c)/(lamda_(2))=(3xx10^(8))/(4047xx10^(-10))=7.412xx10^(14)Hz` `v_(3)=(c)/(lamda_(3))=(3xx10^(8))/(4358xx10^(-10))=6.883xx10^(14)Hz` `v_(4)=(c)/(lamda_(4))=(3xx10^(8))/(5461xx10^(-10))=5.493xx10^(14)Hz` and `v_(5)=(c)/(lamda_(5))=(3xx10^(8))/(6907xx10^(-10))=4.343xx10^(14)Hz` From the given data, we plot the graph, which comes out to be as shown in figure. first FOUR points lie on a straight line but the FIFTH point does not lie on straight line. it means that `v_(5) lt v_(0)`. (a) A careful measurement shows that slope of straight line graph is `4.15xx10^(-15)VS` `therefore (h)/(e)=4.15xx10^(-15)=exx4.15xx10^(-15)=1.6xx10^(-19)xx4.15xx10^(-15)=6.64xx10^(-34)Js` (b) From graph intercept along v-axis is found to be `5xx10^(14)Hz`. `therefore`Threshold frequency `v_(0)=5xx10^(14)Hz` `therefore`Work function `phi_(0)=6.64xx10^(-34)xx5xx10^(14)J=(6.64xx10^(-34)xx5xx10^(14))/(1.6xx10^(-19))eV` `=2.075eV=2.1eV`. |
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