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## GR0877 Problem 41

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (B) According to Einstein’s photoelectric equation: $h \nu = \varphi + K_{max}$, where $K_{max}$ is the maximum kinetic energy of the ejected electron. $K_{max} = eV$ , where $V$ is the stopping potential. From these two equations: $V = \frac{h}{e} \nu - \varphi/e$ and the slope is $\frac{h}{e}$.

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1. 12.10.2011 at 07:48

Unit analysis MIGHT be quicker: we want slope=Volts/Hertz=Volts*seconds, and we have h=Joules*seconds, e=Coloumbs, and phi (work function)=Joules.
Since seconds are already in the numerator of h, let’s start there where Joules*seconds=Farads*Volts^2*seconds because U (energy in Joules) =(CV^2)/2 (Farads*Volts^2). Also, Q=CV, or C=Q/V, so Farads=Coloumbs/Volts. Plug that back into J*s to get the relevant units of h for this problem: Joules*seconds=Farads*(Volts^2)*seconds=(C/V)*(V^2)*s=C*V*s. Now without using anything besides h (all I’ve done so far is rearrange the units of h) we have arrived at our original goal with an extra Coloumbs on top. Divide h by e (the electron charge in units of Coloumb’s) to get rid of it, this is the answer: h/e, choice B.

2. 20.10.2015 at 21:52

I am just adding on to the dimensional analysis comment.

We have established that the slope = Volts/Hertz = Volts * seconds.

e = coulombs, Work function = Joules,

where as h = Joules * second. We can use W = Q * V ie Joule = Coulomb * Volts.
Hence h = Volts * Coulomb * second.

h has an additional Coulomb as compared to the slope which we require. We can just divide it by e to get Volts * second. ie h/e or option B.