Chemistry Electromagnetic Radiation and its Dual Behaviour, Planck's Quantum Theory, Photoelectric Effect, Atomic Spectra and Bohr's Model for H-atom
Click for Only Video

### Topics to be covered :

This lecture covers following topics :

● Electromagnetic Radiation and its Characteristics
● Planck's Quantum Theory
● Black Body Radiation
● Photoelectric Effect
● Observation
● Condition for the Ejection of Electrons

These are the radiations which do not need any medium for propagation e.g. ultraviolet, infrared etc.

James Maxwell (1870) was the first to explain about the interaction between the charged bodies and the behaviour of electrical and magnetic fields on macroscopic level.

An electromagnetic radiation is generated by oscillations of a charged body in a magnetic field or a magnet in an electric field. Maxwell was the first to tell that light waves are associated with oscillating electric and magnetic character.

### Characteristics of Electromagnetic Radiations :

(i) All electromagnetic radiations travel with velocity of light.

(ii) The oscillating electric and magnetic fields are perpendicular to each other and both are perpendicular to the direction in which the wave is travelling(Fig2.6).

(iii) According to Maxwell theory, energy of EM waves is proportional to the square of amplitude.

(iv) Unlike sound waves or water waves, electromagnetic waves do not require medium and can move in vacuum.

(v) There are many types of EM radiations which differ from one another in wavelength. This is called EM spectrum(Fig2.7).

### These radiations are characterised by :

(i) Frequency(nu) : The number of waves which pass through a point in one second. Its unit is s^(-1) or Hertz.

(ii) Wavenumber(barnu) : This is reciprocal of wavelength and defined as the number of wavelenghts per unit length. Its unit is m^(-1) or cm^(-1).

Frequency, wavelength and velocity of light are related as

c = nu lambda
Q 2662123935

The Vividh Bharati station of All India Radio, Delhi, broadcasts on a frequency of 1,368 kHz (kilo hertz). Calculate the
wavelength of the electromagnetic radiation emitted by transmitter. Which part of the electromagnetic spectrum does it belong to?

Solution:

The wavelength lamda is equal to c/nu where c is the speed of electromagnetic radiation in vacuum and nu is the

frequency. Substituting the given values,
we have lamda = c/nu

 = (3.00xx10^8 m)/(1368 kHz)

 = (3.00xx10^8 m)/(1368xx10^3)

 = 219.3 m

This is a characteristic radio wave wavelength.
Q 2612134030

Calculate (a) wave number and (b) frequency of yellow radiation having wavelength 5800 A^0.

Solution:

(a) Calculation of wave number  (barν )

lamda = 5800A^0

barnu = 1/lamda = 1/(5800)

 = 1.724

(b) Calculation of the frequency (nu)

nu = c/lamda = (3xx10^8)/(5800)

### Planck's Quantum Theory :

Wave nature of EM radiations explain the phenomenon such as : -

(i) reflection (ii) refraction (iii) interference (iv) diffraction (v) polarisation

but this could not explain the following phenomenon

(ii) photoelectric effect
(iii) variation of heat capacity of soilds as a function's of temperature.
(iv) line spectra of atoms with special reference to hydrogen.

The above experimental result cannot be explain on the basis of wave theory of light. Planck suggested that atoms and molecules could emit (or absorb) energy only in discrete quantities and not in a continuous manner. Planck gave the name quantum to the smallest quantity of energy that can be emitted or absorbed in the form of EM radiations.

The energy of a quantum of radiation is proportional to its frequency(nu) and is expressed by

E = h nu

h is planck's constant.

h = 6.626 xx 10^(-34) J-s

Q 2632134032

Calculate energy of one mole of photons of radiation whose frequency is 5 xx10^(14)Hz.

Solution:

Energy (E) of one photon is given by the expression

E = hv

h = 6.626xx10^(-34)J s

nu = 5xx10^(14) s^(-1)  (given)

E = (6.626xx10^(-34) J s)xx(5xx10^(14) s^(-1) )

 = 3.313 xx 10^(-19)J

Energy of one mole of photons

 = (3.313xx10^(-19)J) xx (6.022xx10^(23) mol^(-1))

 = 199.51 kJ mol^(-1)
Q 2652134034

A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per second by the bulb

Solution:

Power of the bulb = 100 watt

 = 100 Js^(-1)

Energy of one photon E = hnu = (hc)/lamda

 = (6.626xx10^(-34)xx3.00xx^8)/(400)

 = 4.969xx10^(-19)

So, Number of photons emitted

=  (100 J s^(-1))/(4.969xx10^(-19) J)
Q 2682834737

The energy required to break one mole of Cl-Cl bonds in Cl_2 is 243 kJ mol^(-1)
The largest wavelength of light capable of breaking a single Cl-Cl bond is

(A)

700 nm

(B)

494 nm

(C)

596 nm

(D)

640 nm

Solution:

Energy required for one Cl_2 molecule = (242xx10^3 )/(N_A) J

E = (hc)/lamda

or lamda = (hc)/E = (6.626xx10^(-34) xx3xx10^8xx6.02xx10^(23))/(242xx10^3)

 = 494xx10^(-9) m = 494 nm
Correct Answer is => (B) 494 nm

=> When an iron rod is treated in a furnace, it first turns due to rod and then progressively becomes more and more red as the tempreature increases.

=> On further treating , the emitted radiation becomes write and then becomes blue as the temprature becomes very high.

=> It means frequency of emitted radiation goes from a lower frequency (red colour) to a higher frequency (blue colour) as the temprature increases.

The ideal body which emits and absorbs radiation of all frequencies is called a black body and the radiation emitted by such a body
is called a black body radiation. The explanation of the phenomenon of the black body radiation was given by Max Planck in 1990.

When soilds are heated they emit radiation over a wide range of wavelengths.

The exact frequency distribution of the emitted radiation ( i.e, intensity vs. frequency curve of the radiation ) from a black body depends only on its temprature (Fig.2.8).

At a given temperature, intensity of radiation emitted increases with decrease of wavelength, reaches a maximum value at a given wavelength

### Photoelectric Effect :

H. Hertz performed a very interesting experiment (1887) in which electrons (or electric current) were ejected when certain metals (for example K, Rb, Cs, etc.) were exposed to a beam of light as shown in fig. 2.9. The phenomenon is called photoelectric effect.
Q 2602134038

When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a kinetic energy of 1.68 ×10^5 J mol^(–1). What is the minimum energy needed to remove an electron from sodium? What is the maximum wavelength that will cause a photoelectron to be emitted ?

Solution:

The energy (E) of a 300 nm photon is given by

hnu = (hc)/lamda

 = (6.626xx10^(-34)xx3.00xx10^8)/(300xx10^(-9))

 = 6.626xx10^(-19)J

The energy of one mole of photons

 = 6.626xx10^(-19) J xx 6.022xx10^(23) mol^(-1)

 = 3.99xx10^5 J mol^(-1)
The minimum energy needed to remove a mole of electrons from sodium(hnu_o = hnu -1/2 mv^2)

= (3.99-1.68)xx10^5 J mol^(-1)

 = 2.31xx10^5 J mol^(-1)

Now, the minimum energy for one electron

 = (2.31xx10^5 )/(6.022xx10^(23))

 = 3.84xx10^(-19)J

This corresponds to the wavelength

therefore lamda = (hc)/E

 = (6.626xx10^(-34)xx3.00xx10^8)/(3.84xx10^(-19)J)

 = 517nm
(This corresponds to green light)
Q 2632234132

The threshold frequency nu_0 for a metal is 7.0xx10^(14)s^(-1). Calculate the kinetic energy of an electron emitted when radiation of frequency ν = 1.0xx10^(15) s^(-1) hits the metal

Solution:

According to Einstein’s equation

Kinetic energy  = 1/2 m_e v^2 = h(nu-nu_0)

 = (6.626xx10^(-34) Js) xx(1.0xx10^(15) s^(-1) -7.0xx10^(14) s^(-1))

 = (6.626xx10^(-34)xxJs) xx (10.0xx10^(14) s^(-1) -7.0xx10^(14) s^(-1))

 = (6.626xx10^(-34)Js) xx (3.0xx10^(14) s^(-1)

 = 1.988xx10^(-19) J

### Observation :

(i) there is no time gap between the striking of light beam and ejection of electron from the metal surface.

(ii) no. of electrons ejected prop intensity or brightness of light

(iii) the emission of photoelectron takes place only when the frequency of the incident radiation is above a critical value, characteristic of metal. This critical value of frequency is known as threshold frequency.

(iv) maximum K.E. of photoelectrons prop frequency of incident radiation.

### Condition for the Ejection of Electrons :

Case I : - when  nu < nu_o

No electron is ejected.

Case II : - when  nu = nu_o

Photoelectrons with zero kinetic energy will be formed .

Case III : - when  nu > nu_o

Photoelectrons with certain kinetic energy will be ejected.

nu = frequency of incident light, nu_0 = threshold fequency

The kinetic energy of ejected electron is given by

h nu = h nu_0 + 1/2 m_e v^2

h nu_0 is also called work function.

 m_e =>  mass of electron.

 v => velocity of ejected electron.