Physics · Dual Nature Of Radiation And Matter · NEET
Start from Einstein's equation: eV0 = h(nu) - phi. Divide both sides by e: V0 = (h/e)nu - (phi/e). This is in the form y = mx + c, which is the equation of a straight line. Here y = V0, x = nu, slope m = h/e, and intercept c = -(phi/e). Because h, e and phi are all constants for a given metal, V0 rises straight up as nu increases. That is why the plot is a clean straight line, not a curve.
The slope is h/e, where h is Planck's constant (6.63 x 10^-34 J s) and e is the electron charge (1.6 x 10^-19 C). So slope = 6.63e-34 / 1.6e-19 = 4.14 x 10^-15 volt-second. This is a UNIVERSAL constant. It does not depend on the metal, the intensity, or the light source. This is the key idea used to find Planck's constant from the graph.
No. The slope stays h/e for every metal, so all the lines are PARALLEL. What changes from metal to metal is the position of the line: a metal with a higher work function (phi) has a larger threshold frequency nu0, so its line is shifted to the RIGHT and starts further along the frequency axis. Same slope, different starting point.
The x-intercept (where the line crosses the frequency axis, V0 = 0) is the threshold frequency nu0. At V0 = 0, Einstein's equation gives h(nu0) = phi, so nu0 = phi/h. Below nu0 no electrons are emitted, so the real graph only starts from nu0. To the left of nu0 the line is just a dashed extension used to read the intercept.
Set nu = 0 in V0 = (h/e)nu - (phi/e). You get V0 = -(phi/e), which is negative because phi and e are positive. So the extended line meets the V0 axis BELOW zero at -(phi/e). This point has no physical meaning on its own (you cannot have zero frequency light), but its magnitude gives you the work function: phi = e x (magnitude of y-intercept).
They look almost the same, but the y-axis differs. KE(max) vs nu has slope h and y-intercept -phi. V0 vs nu has slope h/e and y-intercept -phi/e (because V0 = KE(max)/e). Both cross the frequency axis at the SAME threshold frequency nu0. So use slope h for the energy graph and slope h/e for the stopping potential graph.
When two monochromatic lights of frequency nu and nu/2 are incident on a photoelectric metal, their stopping potentials become Vs/2 and Vs respectively. The threshold frequency for this metal is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
V0 = (h/e)nu - (phi/e). It comes from Einstein's photoelectric equation eV0 = h(nu) - phi divided by e. It is a straight line with slope h/e and y-intercept -(phi/e).
The slope equals h/e = 4.14 x 10^-15 V s. Multiply the slope by the electron charge e to get Planck's constant: h = slope x e.
The x-intercept (V0 = 0) is the threshold frequency nu0 = phi/h. The y-intercept is -(phi/e); its magnitude times e gives the work function phi.
Because the slope h/e is the same for all metals. Only the threshold frequency changes, so a higher work function metal shifts its line to the right while keeping the same slope.
The physical line starts from the threshold frequency nu0 on the frequency axis. Below nu0 there is no emission and no stopping potential, so that part is only a dashed extension for reading the y-intercept.
