Q.In a photoelectric experiment carried out with two different metals A and B, the stopping potential (in volts) is measured as a function of the frequency (in Hz) of the incident light. For each metal the measured points lie on a straight line, and the two straight lines are parallel to each other (they have the same slope). The line for metal A lies to the left of the line for metal B. Extrapolated down to zero stopping potential, the line for A meets the frequency axis at a threshold frequency of about Hz, while the line for B meets it at about Hz; each line then rises with the same slope (for instance, line A passes close to the point Hz, V, and line B passes close to the point Hz, V).
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Start your 14-day free trial to unlock the full solution →Both metals give parallel straight lines of versus , so they share the same slope . Metal B's line meets the frequency axis at the higher threshold frequency, so B has the larger work function. The common slope gives J·s for both metals, matching Einstein's constant — confirming his theory.
Concept — Einstein's photoelectric equation
The stopping potential satisfies
which rearranges to
This is a straight line in the plane with slope and a frequency-axis intercept (where ) at the threshold frequency .
(i) Which has the higher work function
The work function is , so a larger threshold frequency (an intercept further to the right on the frequency axis) means a larger work function. Line B meets the axis at Hz while line A meets it at Hz, so
Metal B has the higher work function.
(ii) Value of Planck's constant
The slope of each line is , so .
For material A, using at Hz and V at Hz:
For material B, using at Hz and V at Hz:
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