Question

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 end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used λ1 = 3650 Å, λ2= 4047 Å, λ3= 4358 Å, λ4= 5461 Å, λ5= 6907 Å, The stopping voltages, respectively, were measured to be V01 = 1.28 V, V02 = 0.95 V, V03 = 0.74 V, V04 = 0.16 V, V05 = 0 V Determine the value of Planck’s constant h, the threshold frequency and work function for the material. [Note You will notice that to get h from the data, you will need to know e (which you can take to be 1.6 × 10−19 C). Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of e (from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of h.]

Answer 1

In order to calculate plank's constant (h),  we need to find slope of the graph between cut off voltage and frequency.

so, first of all, let's calculate the frequency (v = c/λ) in each case. in figure shows corresponding stopping potential.

now, draw graph of $V_0$ vs $\nu$. see 2nd figure. it shows that the first four points lie nearly on a straight line which intercepts x -axis at threshold frequency, $\nu_0$ = 5 × 10¹⁴ Hz. The fifth point v = 4.3 × 10¹⁴ < $\nu_0$ so, there is no photoelectric emission and no stopping voltage is required to stop the current.

slope of graph = h/e

or, ∆Vo/∆v = h/e

or, (1.28 -0)/(8.2 - 5) × 10¹⁴ = h/(1.6 × 10^-19)

or, (1.28/3.2) × 10^-14 = h/(1.6 × 10^-19)

or, h = 6.4 × 10^-34 J.s

hence, plank's constant , h = 6.4 × 10^-34 J.s

(b) Threshold frequency, $\nu_0$ = 5 × 10¹⁴ Hz

so, work function = h $\nu_0$

= 6.4 × 10^-34 × 5 × 10¹⁴

= 32 × 10^-20 J

we know, 1 eV = 1.6 × 10^-19 J

so, work function = (32 × 10^-20)/(1.6 × 10^-19) = 2eV