Physics · Dual Nature Of Radiation And Matter · NEET
Start from Einstein's equation: eV0 = hν - φ0. Divide both sides by e to make V0 the subject: V0 = (h/e)ν - (φ0/e). Compare with y = mx + c. The y-axis is V0 and the x-axis is ν, so the slope m = h/e. That is why you must multiply the slope by e to get h: h = e x slope. The slope alone is not h.
Step 1: Pick two clear points on the straight line, (ν1, V01) and (ν2, V02). Step 2: slope = (V02 - V01) / (ν2 - ν1). This has units volt per hertz, which is joule-second per coulomb. Step 3: multiply by the electron charge: h = e x slope = 1.6 x 10^-19 x slope. The answer should come out near 6.6 x 10^-34 J s.
The x-intercept (where V0 = 0) is the threshold frequency ν0. Below this frequency no electrons come out. The y-intercept is at -φ0/e (a negative value on the V0 axis), so the line, if extended back, cuts the V0 axis below zero. From either intercept you can get the work function: φ0 = h ν0, or φ0 = e x (magnitude of the y-intercept).
Yes. The slope is h/e, and both h and e are universal constants. So the V0 vs ν lines for caesium, sodium, potassium are all parallel with the same slope. Only the intercepts differ, because each metal has a different work function φ0 and different threshold frequency ν0. This parallel-lines fact is a favourite NEET point.
Two easy ways. (1) Read the x-intercept ν0, then φ0 = h ν0 (use the h you just found). (2) Read the magnitude of the y-intercept, call it V0-intercept, then φ0 = e x V0-intercept. Both give the same answer. To express φ0 in electron-volts, divide the joule value by 1.6 x 10^-19.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
h = e x slope of the stopping potential versus frequency graph, where e = 1.6 x 10^-19 C. The slope itself equals h/e.
Stopping potential V0 on the y-axis and frequency ν of the incident light on the x-axis. The plot is a straight line described by V0 = (h/e)ν - φ0/e.
Because the slope is h/e, made only of universal constants, so it is identical for all metals. Different metals shift the line up or down through their different work functions, but never change the slope.
About 6.63 x 10^-34 J s. If your graph calculation gives a number close to this, your slope reading is correct.
Yes. Use the x-intercept ν0 with φ0 = h ν0, or use the magnitude of the negative y-intercept with φ0 = e x that intercept.
