ATOMIC PHYSICS

Frank Hertz experiment

Frank and Hertz performed an experiment to proof the existence of energy level within an atom. In the experiment, they measure the energy needed to raise an electron from its ground state to a higher state called excited state.

Diagram/ Set up of experiment:

Description of the setup:

From the diagram above, the tube contains mercury or sodium vapor at very low pressure. A cathode and anode are placed at the opposite sides of the tube. A grid is placed between the cathode and the anode, but closer to the anode. A galvanometer is connected in series with the anode. A voltmeter, potentiometer are connected as shown in the diagram. The anode and the cathode are separately powered

Explanation of the experiment:

The tube contains mercury or sodium vapor at very low pressure. When the circuit is switched on, the cathode become hot and emits electrons by thermionic emission. The emitted electrons accelerate to the anode which is positively biased. As the electrons incident on the anode, current flow in the galvanometer. The grid placed between the cathode and anode is negatively biased to help regulate the amount of electrons that flow to the anode. This in turn helps to regulate the current that flow. Note that the grid is placed near the anode to allow the electrons collide before reaching the anode.

Graph of anode current against anode voltage

In the experiment, different values of anode current are obtained for different values of voltages. These different values of currents are plotted against the different values of voltage and the graph as shown below is obtained.

 

V₁,v₂ and v₃ are critical voltages. At these voltages, the electrons have just enough kinetic energy to raise

the internal energy of the sodium atom by collision.

PHOTO ELECTRIC EFFECT

Photo electric effect is the emission of electron from the surface of a metal when light of a particular frequency falls on the surface of the metal. The electrons emitted is called photo electron.

The maximum energy of electron emitted is directly proportional to the frequency of the light that incident on the metal surface. It does not depend on the intensity of the incident light.

Photo cell

Diagram of a photo cell:

 EXPLANATION OF WORKING PRINCIPLE OF PHOTO CELL

When ultraviolent light ray of a particular frequency incident on the cathode of the cell, electrons are emitted from the surface of the cathode by photo electric emission the electrons accelerate to the positively biased anode and causes current to flow in the circuit.

THRESHOLD FREQUENCY

It is the minimum frequency of incident light that will emit electron from the surface of the metal.

                                                                    

GRAPH OF MAXIMUM ENERGY AGAINST FREQUENCY

the graph is obtained by plotting different values of energy of electron on the vertical axis against different values of frequency on the horizontal axis.

diagram of graph here:
   
                                                                                                             Interpolation of graph
                                                                                          

                                                                                            The slop of the graph represent plank constant.

                                                                                            The intercept oc represent the work function 

                                                                                            of  the metal.

                                                                                            The intercept of₀ is the threshold frequency                                

A straight line graph which cut off the frequency axis at Fo is obtained. Below Fo, no electron is emitted from the surface of the metal. This is because the frequency of the incident light is below the threshold frequency that is required for emission of electron. Increase in the intensity of light increases the number of photo electrons emitted and not the energy and velocity of the electron. When light of frequency that is lower than the threshold frequency falls on the metal surface, the light is absorbed by the metal surface. The absorbed incident light is used to raised the energy level of the surface electron or to overcome the potential barrier of the photo electron.

APPLICATION OF PHOTO CELL

i.It is used in opening and closing of door automatically.  
ii. It is used in automatic switching.

iii. it is used in burglary alarm
Iv. It is used as safety device in paper mill and related industries.

v.   it is used to produce sound of different intensity on film.  
vi. In design of photometer

vii. in cathode ray tube.

LIBRATION OF ELECTRON FROM METAL SURFACE 

 

diagram here

The incident light or photon incidenting on the metal surface has energy hf. This energy hf consist of two parts. One part is the work function w and the other part is the maximum kinetic energy K.Emax.

The work function Wo is used to separate the electron from the atom ( that is to liberate the electron from the nuclear force that keep the electron in the shell.

The k.E max is used to give the electron a kinetic energy ( K.E = ½mv² ) so that the electron can escape from the metal surface. V is the velocity of the emitted electron.

PHOTON:

Photon is packet light energy or particles.

Work FUNCTION (W₀)

Work function is the minimum amount of energy that is needed to knock out / pullout electron from the surface  of a metal. Work function varies from one metal to another.

Value of work function:

The value of work function is,  w₀ = 2.0 eV.
Recall that 1.0 ev = 1.602 x 10̄  ̄¹⁹ J

Therefore, if w₀  = 2.0 eV
Then, 2.0 eV = 2.0 x 1.602 x 10̄  ̄¹⁹ J = 3.2 x 10̄  ̄¹⁹ J for sodium.
Therefore we can say that,
Work function, Wo for Sodium = 3.2 x 10̄  ̄¹⁹ J.

For caesium, Wo₀ = 1.35 eV = 1.35 x 1.602 x 10̄  ̄¹⁹ J = 2.2 x 10̄  ̄¹⁹ J

FORMULA FOR CALCULATING WORK FUNCTION (W₀)

Remember that we define work function (Wo) to be energy. Therefore,

Wo₀ = energy.

Also remember that the energy require to emit electrons from the surface of a metal is

E = hf₀.

Therefore,
 work function(Wo)  = Energy require to emit electron from the surface of a metal

Therefore,

w₀ = E = hf₀.

Where f₀ is the threshold frequency / minimum frequency of light that is required to emit electron from the surface of a metal.

MAXIMUM ENERGY OF EMITTED ELECTRON

Maximum kinetic energy is needed to emit Electrons from the surface of a metal.

Therefore, the formula for calculating the maximum K.E is given by the formula:

 Max K.E = hf - w₀.  

Remember that the formula for K.E = ½mv².

Therefore,          

Max K.E = hf - w₀.

                        ½mv² = hf - w₀.

Recall that frequency = velocity / wavelength i.e f = v / ʎ.  

½mv² = h(v / ʎ ) - w₀.;    
½mv² = h(v / ʎ ) - hf₀

 Note that Wo₀ = hf₀.
Therefore,

½mv² = hf  - hf₀.

 Let us factorize h. then,

         ½mv² = h(f  - f₀).

Where h is Planck constant; ʎ is the wavelength; v is the velocity; m is mass of electron; f is frequency of incident light; f₀ is threshold frequency.

Also, note that electron volt (eV) is the kinetic energy pocess by an electron when the electron is acted upon by a potential difference of 1 volt.

Therefore, we can also say that electron volt = kinetic energy. i.e eV = K.E. therefore in some questions where value of electron volt are given, you take the eV to be the kinetic energy. eV = K.E,

Therefore,  eV =  hf. - w₀.  (Also not that eV = ½mv² )

EINSTEIN ENERGY EQUATION

The formula we obtained above is called the Einstein equation.

Note:
take your time to master all the formulae. By so doing, you will know which of the formula to use when solving problems, base on the parameters that are given in the questions.

Worked examples:

A certain metal has a work function of 2.5 eV. Calculate: i. the threshold frequency ii. Max. K.E of librated electron when the metal is illuminated by light of wavelength 3.5 x 10  ̄⁵ Cm, h = 6.6 x 10  ̄³⁴ Js

C = 3.0 x 10 ⁸ ms  ̄¹, 1 eV = 1.6 x 10   ̄¹⁹ J.

Solution:

Step I:

Let us write down all the parameters given in the question:

Wavelength,  ʎ = 3.5 x 10  ̄⁵ Cm, planck constant h = 6.6 x 10  ̄³⁴ Js

Velocity C = 3.0 x 10 ⁸ ms  ̄¹, 1 eV = 1.6 x 10   ̄¹⁹ J.

 w₀ = 2.5 eV. Change to energy in joul, then w = 2.5 x 1.6 x 10   ̄¹⁹ J = 4.0 x 10   ̄¹⁹ J

 step ii :

Let us write down all the formulae that are involved under this topic question

Formulae:

1.       w₀ = hf₀  ( E = hf₀ ) remember W and E are the same i.e energy

2.       Max K.E = hf - w₀  or   ½mv² =  hf - w₀. or ½mv² = hf - hf₀ or ½mv² = h(v / ʎ ) - hf₀

 (Remember that w₀ = hf₀ )

3.       eV =  hf. - w₀  ( remember that w₀ = hf₀ )

4.       eV = ½mv²

step iii:

use the formula that contains what you want to calculate and two or more of the parameters given in the question.
Therefore let us use formula 1 because it contains what we are to calculate and two parameters that are given in the question.

Therefore,

w₀ = hf₀.  Let us substitute for the parameters. Then we will get,

2.5 x 1.6 x 10   ̄¹⁹ = 6.6 x 10  ̄³⁴ x f₀.

4.0 x 10   ̄¹⁹  = 6.6 x 10  ̄³⁴ x f₀. (Then make f₀ the subject.)

4.0 x 10   ̄¹⁹ / 6.6 x 10  ̄³⁴ = f₀                     

f₀ = 0.606 x 10 ¹⁵ Hertz.

Maximum energy:

Max K.E = hf - w₀  (f = v / ʎ.  Replace f with v / ʎ )

Therefore,

Max K.E = hf - w₀  
 Max K.E = h(v / ʎ) - w₀ . Then substitute for the parameters.

Max K.E = 6.6 x 10  ̄³⁴ x (6.6 x 10  ̄³⁴/3.5 x 10  ̄⁵) - 4.0 x 10  ̄¹⁹

                Max K.E = 6.6 x 10  ̄³⁴ x 1.89 x 10  ̄²⁹ - 4.0 x 10   ̄¹⁹

                Max K.E = 12.474 10  ̄⁶³ - 4.0 x 10  ̄¹⁹
( Note: change the numbers to ordinary forms, subtract them and change back to standard form).

Max K.E =

Calculate the energy in joule of a light photon of wavelength 7000 Armstrong (Å).
1(Å) = 10  ̄¹⁰m ,

C = 3.0 x 10 ⁸ ms  ̄¹,  h = 6.6 x 10  ̄³⁴ Js

Solution:

Step i.

Write down all the parameters given in the question.

Wavelength ʎ =                7000 Armstrong (Å). Change it to meter. 1(Å) = 10  ̄¹⁰m. Then, 7000 Å will be

                7000 Å = 7000 x 10  ̄¹⁰m ,  C = 3.0 x 10 ⁸ ms  ̄¹,  h = 6.6 x 10  ̄³⁴ Js

Step ii.

Let us write down all the formulae that are involved under this topic question

Formulae:

1.       w₀ = hf₀  ( E = hf₀ ) remember W and E are the same i.e energy

2.       Max K.E = hf - w₀ or   ½mv² =  hf - w₀. or ½mv² = hf - hf₀ or ½mv² = h(v / ʎ ) - hf₀

 (Remember that w₀ = hf₀ )

3.       eV =  hf. - w₀  ( remember that w₀ = hf₀ )

4.       eV = ½mv²

Note:

In subsequent worked examples, I will not state step ii. Rather I will just state the formula for solving the problem. I want to believe that by now, you are familiar with step ii.

Step iii:

                            Formula for calculation:  w₀ = E = hf.

Note at times f₀ = f. you use f when the threshold frequency is not given. Let us substitute for the parameters. Then, we will get,

                            E = hf.  E = h x (v / ʎ ).if we substitute for the parameters,  Then

E = 6.6 x 10  ̄³⁴ x ( 3.0 x 10 ⁸ /  7000 x 10  ̄¹⁰ ). E =6.6 x 10  ̄³⁴ x 4.29 x 10  ̄¹⁴.  E = 28.314 x 10  ̄²⁰Joules.

Worked example:

The gr0ound state of hydrogen is -13.6 eV and the first excited state is -3.4 eV.  The state in which an electron is completely free from the atom is zero energy. Calculate the wavelength of radiation.

(h = 6.6 x 10  ̄³⁴Js, 1 eV = 1.6 x 10   ̄¹⁹ J.

Solution:

                 E₀ = -13.6 eV = 13.6 x 1.6 x 10   ̄¹⁹ J = 21.76 x 10   ̄¹⁹ J

                 En = -3.4 eV = -3.4 x x 1.6 x 10   ̄¹⁹ J = - 5.44 x 10   ̄¹⁹J

                 1 eV = 1.6 x 10   ̄¹⁹ J, h = 6.6 x 10  ̄³⁴ J

State the formula for calculation:

From the question, note that the electron will return from the excited state to its ground state when the disturbance is removed. In this process, the electron will give out the excess energy it received. This energy is the energy u will use for the calculation. Then, Energy = En - E₀ .

               Formula:     En - E₀ = hf .  Recall that f = v / ʎ. then,                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                 

                                    En - E₀ = h x ( v / ʎ ). Let us substitute for the parameters.

                                    -13.6 – (- 3.4 ) eV = 6.6 x 10  ̄³⁴  x (3.0 x 10 ⁸/ ʎ)

                                     (-13.6 + 3.4) eV = 6.6 x 10  ̄³⁴  x (3.0 x 10 ⁸/ ʎ).

                                     -10.2 eV = 6.6 x 10  ̄³⁴ x (3.0 x 10 ⁸/ ʎ). Make ʎ the subject. Then,

                                     ʎ = 6.6 x 10  ̄³⁴  x 3.0 x 10 ⁸ / ( -13.6 + 3.4) eV. Change eV to joule. Then,

                                     ʎ = 6.6 x 10  ̄³⁴  x 3.0 x 10 ⁸ / -10.2 eV. Change eV to joule. Then,

                                     ʎ = 6.6 x 10  ̄³⁴  x 3.0 x 10 ⁸ / 10.2 x 1.6 x 10   ̄¹⁹ ( ignore the minus sign)

                                     ʎ = 19.8 x 10  ̄²⁶ / 16.3 x 10   ̄¹⁹

                                     ʎ = 1.21 x 10   ̄⁷ meter.

Supposing you are ask to calculate the frequency, then we can use the formula, f = v / ʎ. If we substitute for the parameters,

                    Then, f = v / ʎ = 3.0 x 10 ⁸ / 1.21 x 10   ̄⁷. F = 2.48 x 10   ̄¹⁵ Hertz

More worked examples: