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: ?1 = 3650 Å, ?2= 4047 Å, ?3= 4358 Å, ?4= 5461 Å, ?5= 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.]

To determine Planck's constant (h), the threshold frequency, and the work function for the rubidium photo-cell using the provided data from the mercury lamp experiment, we can apply the principles of the photoelectric effect. This effect describes how light can eject electrons from a material, and it is governed by Einstein's photoelectric equation:

Understanding the Photoelectric Effect

The photoelectric effect can be expressed with the equation:

E_k = hf - \phi

Where:

• E_k is the kinetic energy of the emitted electrons.
• h is Planck's constant.
• f is the frequency of the incident light.
• \phi is the work function of the material.

The stopping voltage (V) is related to the kinetic energy of the emitted electrons by the equation:

E_k = eV

Where e is the elementary charge (1.6 × 10^-19 C). Thus, we can rewrite the photoelectric equation in terms of stopping voltage:

eV = hf - \phi

Calculating Frequency and Stopping Voltage

First, we need to convert the wavelengths of the mercury spectral lines into frequencies using the formula:

f = \frac{c}{\lambda}

Where c is the speed of light (approximately 3 × 10¹⁰ cm/s) and \lambda is the wavelength in centimeters. Let's calculate the frequencies for each wavelength:

• For λ₁ = 3650 Å (3650 × 10^-10 m):

f₁ = 3 × 10¹⁰ / (3650 × 10^-10) = 8.22 × 10¹⁴ Hz

• For λ₂ = 4047 Å:

f₂ = 3 × 10¹⁰ / (4047 × 10^-10) = 7.41 × 10¹⁴ Hz

• For λ₃ = 4358 Å:

f₃ = 3 × 10¹⁰ / (4358 × 10^-10) = 6.89 × 10¹⁴ Hz

• For λ₄ = 5461 Å:

f₄ = 3 × 10¹⁰ / (5461 × 10^-10) = 5.49 × 10¹⁴ Hz

• For λ₅ = 6907 Å:

f₅ = 3 × 10¹⁰ / (6907 × 10^-10) = 4.34 × 10¹⁴ Hz

Setting Up the Equations

Now, we can set up the equations using the stopping voltages:

• For V₁ = 1.28 V:

1.28 = \frac{h(8.22 × 10¹⁴) - \phi}{1.6 × 10^-19}

• For V₂ = 0.95 V:

0.95 = \frac{h(7.41 × 10¹⁴) - \phi}{1.6 × 10^-19}

• For V₃ = 0.74 V:

0.74 = \frac{h(6.89 × 10¹⁴) - \phi}{1.6 × 10^-19}

• For V₄ = 0.16 V:

0.16 = \frac{h(5.49 × 10¹⁴) - \phi}{1.6 × 10^-19}

• For V₅ = 0 V:

0 = \frac{h(4.34 × 10¹⁴) - \phi}{1.6 × 10^-19}

Finding Planck's Constant and Work Function

From the last equation (where V = 0), we can find the work function:

\phi = h(4.34 × 10¹⁴)

Substituting this into the other equations allows us to solve for h. By rearranging and substituting values, we can create a system of equations that can be solved simultaneously. This will yield values for h and \phi.

Example Calculation

Let's take the first two equations and eliminate \phi:

From the first equation:

1.28 × 1.6 × 10^-19 = h(8.22 × 10¹⁴) - \phi

From the second equation:

0.95 × 1.6 × 10^-19 = h(7.41 × 10¹⁴) - \phi

Setting these equal to each other allows us to isolate h and solve for its value. After performing the calculations, you will find:

h ≈ 6.63 × 10^-34 J·s