# Dual Behavior of Electromagnetic Radiation And Line Spectrum of Huydrogen

## Dual Behavior of Electromagnetic Radiation:

Electromagnetic radiation posses both wave as well as particle nature i. e. dual nature.

## SPECTRUM:-

When white light is passed through prism, it splits into band of seven colors called spectrum.
VIBGYOR (V) = Shortest wavelength 400nm; R = Longest wavelength 450 nm

### Continuous spectrum:-

When white light is analyzed by passing through prism, it slits into seven clowns. These clowns are so continuous that each of them merges into the next. Hence, this spectrum known as continuous spectrum.

### Emission Spectrum:-

It is noticed when radiations emitted from source are passed through a prism & received on photographic plate.

### Absorption Spectrum:-

Absorption spectrum is the spectrum obtained when radiation is passed through a sample of material. The sample absorbs radiation of certain wavelengths. The wavelengths which are absorbed are missing and come as dark lines.

### Line Spectrum:-

When vapors of volatile substance is passed through the Bunsen burner flame and then analyzed with the help of spectroscope, such specific coloured lines appear on the photographic plates (different for different substances).
The study of emission or absorption spectra is referred as spectroscopy

## LINE SPECTRUM OF HYDROGEN:-

When hydrogen gas present on discharge tube passed through high voltage & low pressure, the radiations emitted passed through spectroscope and spectrum is obtained on photographic plate. Hydrogen spectrum contains several lines known as series. The series present in hydrogen spectrum are:-
 Series n1 n2 Spectral region Lyman 1 2,3,4,.. Ultraviolet Balmer 2 3,4,5…. Visible Paschen 3 4,5,6,7.. Infrared Brackett 4 5,6,7… Infrared Pfund 5 6,7,8… Infrared

Balmer showed that wave no. of spectral lines in visible region is given by
$= 109677( \frac {1}{2^2} - \frac {1}{n^2})$  cm-1
where n is an integer equal to or greater than 3 (i.e., n = 3,4,5,....)
Wave no. for any two series is obtained by
$= R( \frac {1}{n_1^2} - \frac {1}{n_2^2})$   cm-1
n1, n2 = level of series
The value R = 109,677 cm-1 is called the Rydberg constant for hydrogen.

Question 1
A proton of wavelength 400 nm strikes metal surface. The electrons are ejected with velocity 5.85 × 105 m/s. Calculate min. energy required to remove electron from metal surface. (Mass of electron = 9.1 × 10-31 kg)
Solution
Given:-
λ = 400 nm
Velocity = 5.85 × 105 m/s
C = νλ

3 × 108 m/s = ν. 400 × 10(-9) m
or
ν= 7.5 × 10(14) m
Now E = hν
= 6 × 10-34 × 7.5 × 1014
= 45 ×10-20 J
Kinetic Energy of Electron =1/2 mv2
=.5 × 9.1 × 10-31 × 5.85 × 105 × 5.85 × 105
=15.57 ×10-20 J
Min Energy = 45 ×10-20 J - 15.57 ×10-20 J = 29.43 ×10-20 J
Question 2
Calculate the wavelength of the spectral line in the spectrum of hydrogen tom when n = 3 in Balmer formula.
Solution
$\frac {1}{\lambda } = R (\frac {1}{(n_1^2}- \frac {1}{n_2^2 })$
For Balmer series n1=2
$\frac {1}{\lambda } = R (\frac {1}{(2^2}- \frac {1}{n_2^2 })$
Now for n1=3
$\frac {1}{\lambda } = R (\frac {1}{(2^2}- \frac {1}{3^2 })$
λ= 36/5 R = 6.792 × 10-5 cm
Question 3
In hydrogen atom the electron jumps from 3rd orbit to 1st orbit. Find out fequency & wavelenght of spectrum line.
Solution
$\frac {1}{\lambda } = R (\frac {1}{(n_1^2}- \frac {1}{n_2^2 })$
$\frac {1}{\lambda } = R (\frac {1}{(1^2}- \frac {1}{3^2 })$
$\frac {1}{\lambda }=109677 (1- \frac {1}{9})$
λ= = 9/8 × 1/(1.09 × 105 )
= 1.02 × 10-5 cm.= 102 × 10-5 m
Now, C = νλ

ν=(3 × 108)/(102 × 10-5 )
=(3 × 1013)/102
= 2.67 × 1011 s-1
Question 4
Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol-1.
Solution
From E= hv or hc/ ?, we have ? = 242nm or 242 x 10-9m
C = 3 × 108 m/s
h = 6.62 × 10-34 Js
Now putting these values in the equation we get
E= (6.62 × 10-34 Js × 3 × 108 m/s) / (242 × -9m)
= 0.0821 × 10-17 J/atom
Or
E = (0.0821 x 10-17 ) / (1000 x 6.02 x 1023)
= 494 KJ/mol
Question 5
What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?
Solution
λ= ?
n = 4, n = 2
$\frac {1}{\lambda } = R (\frac {1}{(n_1^2}- \frac {1}{n_2^2 })$
$\frac {1}{\lambda } = R (\frac {1}{(2^2}- \frac {1}{4^2 })$
? = 16 / 109678 × 3 cm = 486 nm
Question 6
Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.
Solution
$\frac {1}{\lambda } = R (\frac {1}{(n_1^2}- \frac {1}{n_2^2 })$
For Balmer series n1=2
$=\frac {1}{\lambda } = R (\frac {1}{(2^2}- \frac {1}{n_2^2 })$
To get maximum wavelength, wave no. must be least. For wave no. to be least n2 should be minimum
Therefore n2 = 3
$= R (\frac {1}{(2^2}- \frac {1}{n_2^2 })$
= 1.09 × 105 × 5/36 = 1.5 × 104 cm-1
Question 7
Arrange the following types of radition in increasing order of their wavelength
a) From microwave oven
b) AMBER light from traffic signal