**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:

*λ*1 = 3650 Å, *λ*2= 4047 Å, *λ*3= 4358 Å, *λ*4= 5461 Å, *λ*5= 6907 Å,

The stopping voltages, respectively, were measured to be:

*V*01 = 1.28 V, *V*02 = 0.95 V, *V*03 = 0.74 V, *V*04 = 0.16 V, *V*05 = 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*.]

**Solution:**

Einstein’s photoelectric equation is given as:

$e V_{0}=h v-\phi_{0}$

$V_{0}=\frac{h}{e} v-\frac{\phi_{0}}{e}$ ....(1)

Where,

*V*0 = Stopping potential

*h* = Planck’s constant

*e* = Charge on an electron

ν = Frequency of radiation

$\phi_{0}=$ Work function of a material

It can be concluded from equation (1) that potential *V*0 is directly proportional to frequency *ν*.

Frequency is also given by the relation:

$v=\frac{\text { Speed of light }(c)}{\text { Wavelength }(\lambda)}$

This relation can be used to obtain the frequencies of the various lines of the given wavelengths.

$v_{1}=\frac{c}{\lambda_{1}}=\frac{3 \times 10^{8}}{3650 \times 10^{-10}}=8.219 \times 10^{14} \mathrm{~Hz}$

$v_{2}=\frac{c}{\lambda_{2}}=\frac{3 \times 10^{8}}{4047 \times 10^{-10}}=7.412 \times 10^{14} \mathrm{~Hz}$

$v_{3}=\frac{c}{\lambda_{3}}=\frac{3 \times 10^{8}}{4358 \times 10^{-10}}=6.884 \times 10^{14} \mathrm{~Hz}$

$v_{4}=\frac{c}{\lambda_{4}}=\frac{3 \times 10^{8}}{5461 \times 10^{-10}}=5.493 \times 10^{14} \mathrm{~Hz}$

$v_{5}=\frac{c}{\lambda_{5}}=\frac{3 \times 10^{8}}{6907 \times 10^{-10}}=4.343 \times 10^{14} \mathrm{~Hz}$

The given quantities can be listed in tabular form as:

The following figure shows a graph between *ν*and *V*0.

It can be observed that the obtained curve is a straight line. It intersects the *ν*-axis at 5 × 1014 Hz, which is the threshold frequency (*ν*0) of the material. Point D corresponds to a frequency less than the threshold frequency. Hence, there is no photoelectric emission for the *λ*5 line, and therefore, no stopping voltage is required to stop the current.

Slope of the straight line $=\frac{\mathrm{AB}}{\mathrm{CB}}=\frac{1.28-0.16}{(8.214-5.493) \times 10^{14}}$

From equation (1), the slope $\frac{h}{e}$ can be written as:

$\frac{h}{e}=\frac{1.28-0.16}{(8.214-5.493) \times 10^{14}}$

$\therefore h=\frac{1.12 \times 1.6 \times 10^{-19}}{2.726 \times 10^{14}}$

$=6.573 \times 10^{-34} \mathrm{Js}$

The work function of the metal is given as:

$\phi_{0}=h \mathrm{v}_{0}$

= 6.573 × 10−34 × 5 × 1014

= 3.286 × 10−19 J

$=\frac{3.286 \times 10^{-19}}{1.6 \times 10^{-18}}=2.054 \mathrm{eV}$

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