Einstein’s Equation

Einstein proposed that light waves consist of packets of energyEnergy is the ability to do work. Energy exists in several forms such as heat, kinetic or mechanical energy, light, potential energy, and electrical energy. Units of Energy: The SI unit... More called quanta or photons and each photon has an energy, hf.

• h = Plank’s constant (6.63 × 10^-34 Js)
• f is the frequency of the light

That is:

\(\scriptsize E = hf \)

but f = \( \frac{c}{\lambda} \)

∴ E = \( \frac{hc}{\lambda} \)

The incident light must produce enough energy for electrons to overcome the electrostatic force of the metal. The work done by an incident light to remove electrons from a metal is called the work function (W₀).

What is Work Function?

Work function (W₀) is a metal property referring to the minimum energy needed to liberate an electron from a metal surface. The symbol of work function is Φ.

Different metals have different work functions because It depends on the nature of the metal surface and property of the metal. The electronic structures of different metals vary, meaning the amount of energy required to remove an electron from the surface of a metal is unique to each metal.

Work Function for Some Common Metals:

Metal	Work function Φ (eV)
Na	2.46
Al	4.08
Pb	4.14
Zn	4.31
Fe	4.50
Cu	4.70
Ag	4.73
Pt	6.35

When plotting energy against the frequency of radiation, the slope of the resulting line represents Planck’s constant (h)

Using Einstein’s Relation:

\(\scriptsize hf = W_0 \: +\: \normalsize \frac{1}{2} \scriptsize mv^2\)

• hf represents the total energy content of a single photon of light
• W₀ = Work function of metal

the remainder of the energy is used to give the liberated electron, kinetic energy

i.e. \( \frac{1}{2} \scriptsize mv^2\)

• v = velocity of photo-electron
• m = mass of photo-electron

The work function, W₀, is given by:

W₀ (Φ) = hf₀

• where f₀ is the threshold frequency
• W₀ is the work function
• h is Plank’s constant

∴ hf = hf₀ + \( \frac{1}{2} \scriptsize mv^2\)

∴ hf = hf₀ + E

E = \( \frac{1}{2} \scriptsize mv^2\) ⇒ Kinetic energy of the photoelectrons.

∴ E = hf – hf₀

This is known as Einstein’s photoelectric equation.

Stopping potential, or cut-off potential, is defined as the minimum potential difference required to stop the removal of an electron from a metal surface when the incident light energy is greater than the work potential of the metal on which the incident light is focused.

The work function and stopping potential tend to mean the same thing in terms of the photoelectric effect.

The energy of the ejected electron may be found by determining what potential difference V must be applied to stop its motion.

E = \( \frac{1}{2} \scriptsize mv^2 \scriptsize = eV\)

⇒ \( \frac{1}{2} \scriptsize mv^2\scriptsize = eV\)

• eV is electron Volt.

∴ \(\scriptsize E = eV = hf \: -\: hf_0 \)

The photoelectric effect only occurs if the energy of the incident light is greater than the work function of the material and the frequency of the incident light is greater than the threshold frequency of the metal.

Example 11.4.1:

Calculate the frequency of the photons whose energy is required to eject surface electrons with a K.E. of 3.8 × 10^-17 eV and the work function of the metal is 2.66 × 10^-17 eV.

h = 6.63 × 10^-34 Js, 1 ev = 1.6 × 10^-19 J

Solution

E = hf – hf₀

E = hf – W₀

hf = E + W₀

hf = \(\scriptsize 3.80 \times 10^{-17} \: + \: 2.66 \times 10^ {-17} \)

hf = \(\scriptsize 6.46 \times 10^{-17} \: eV\\ \scriptsize = 6.46 \times 10^{-17} \: \times \: 1.6 \times 10^{-19} \: J \)

∴ f = \( \frac{6.46 \times 10^{-17} \: \times \: 1.6 \times 10^{-19}}{6.63 \times 10 ^{-34}} \)

f = \( \frac{1.0336 \times 10^ {-35}}{6.6 \times 10 ^{-34}} \)

⇒ f = 0.01566 Hz

Example 11.4.2:

The potential difference between the cathodeA cathode is the electrode from which a conventional current leaves the electrolyte. It is the negative part of the cell where reduction takes place. More and target of an x-ray tube is 5.00 × 10⁴ V and the current in the tube is 2.00 × 10^-2 A. Given that only one percent of the total energy supplied is emitted as x-radiation, determine the:

(i) Maximum frequency of the emitted radiation.

(ii) Rate at which heat is removed from the target in order to keep it at a steady temperature.

[planck’s constant, h = 6.63 × 10^-34Js; electronic charge e = 1.60 × 10^-19C]

Solution

• h = 6.63 × 10^-34Js
• e = 1.6 × 10^-19 C
• V = 5.00 × 10⁴ V            
• l = 2.0 × 10^-2 A

(i) From hf = eV

maximum frequency, f = \( \frac{eV}{h} \\ =\frac{1\% \: of \: eV}{h}\\ = \frac{0.001 \: \times \: 1.60\: \times \: 10^{-19} \: \times \: 5.00 \: \times \: 10^4}{6.63 \: \times \: 10^{-34}}\\ = \scriptsize 1.207 \: \times \: 10^7 \: Hz \)

∴ Maximum frequency = \( \scriptsize 1.207 \: \times \: 10^7 \: Hz \)

(ii) Rate of heat energy removal = power, P = IV (100 – 1%) or 99% of heat has to be removed.

P = 99% × I × V

= 0.99 × 2.00 × 10^-2 × 5.00 × 10⁴

= 990 W or 990 Js^-1