- Structure of Atom
- |
- Discovery of Electron
- |
- Discovery of Protons
- |
- Discovery Of Neutron
- |
- Model Of Atoms
- |
- Atomic Terms
- |
- Electromagnetic Theory
- |
- Planck's Quantum Theory
- |
- Photo-electric effect
- |
- Dual Behavior of Electromagnetic Radiation
- |
- Line Spectrum Of Hydrogen
- |
- BOHR MODEL for Hydrogen Atom
- |
- De-Broglie Equations
- |
- Heisenberg Uncertainty Principle
- |
- Quantum Mechanical Model
- |
- Quantum Numbers
- |
- Shapes OF Orbital
- |
- Rules for the filling of electrons in the Orbital

VIBGYOR (V) = Shortest wavelength 400nm; R = Longest wavelength 450 nm

The study of emission or absorption spectra is referred as spectroscopy

Series |
n_{1} |
n_{2} |
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

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

n

The value R = 109,677 cm

A proton of wavelength 400 nm strikes metal surface. The electrons are ejected with velocity 5.85 × 10

Given:-

λ = 400 nm

Velocity = 5.85 × 10

C =

3 × 10

or

ν= 7.5 × 10

Now E = hν

= 6 × 10

= 45 ×10

Kinetic Energy of Electron =1/2 mv

=.5 × 9.1 × 10

=15.57 ×10

Min Energy = 45 ×10

Calculate the wavelength of the spectral line in the spectrum of hydrogen tom when n = 3 in Balmer formula.

$ \frac {1}{\lambda } = R (\frac {1}{(n_1^2}- \frac {1}{n_2^2 })$

For Balmer series n

$ \frac {1}{\lambda } = R (\frac {1}{(2^2}- \frac {1}{n_2^2 })$

Now for n

$ \frac {1}{\lambda } = R (\frac {1}{(2^2}- \frac {1}{3^2 })$

λ= 36/5 R = 6.792 × 10

In hydrogen atom the electron jumps from 3rd orbit to 1st orbit. Find out fequency & wavelenght of spectrum line.

$ \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 × 10

= 1.02 × 10

Now, C =

ν=(3 × 10

=(3 × 10

= 2.67 × 10

Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol

From E= hv or hc/ ?, we have ? = 242nm or 242 x 10

C = 3 × 10

h = 6.62 × 10

Now putting these values in the equation we get

E= (6.62 × 10

= 0.0821 × 10

Or

E = (0.0821 x 10

= 494 KJ/mol

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?

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

Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

$ \frac {1}{\lambda } = R (\frac {1}{(n_1^2}- \frac {1}{n_2^2 })$

For Balmer series n

$

To get maximum wavelength, wave no. must be least. For wave no. to be least n

Therefore n

$

Arrange the following types of radition in increasing order of their wavelength

a) From microwave oven

b) AMBER light from traffic signal

c) Radiation from FM Radio

d) Cosmic rays from outer space

e)X - rays.

Cosmic rays < X-rays < radiation from microwave ovens < amber light < radiation of FM radio

Class 11 Maths Class 11 Physics Class 11 Chemistry

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