`●` Foundation of the Quantum Mechanical Model of the Atom

`●` Dual Behaviour of Matter :

`●` Heisenberg's Uncertainty Principle :

`●` Explanation :

`●` Significance of Uncertainty Principle :

`●` Dual Behaviour of Matter :

`●` Heisenberg's Uncertainty Principle :

`●` Explanation :

`●` Significance of Uncertainty Principle :

Two important developments contributed to this model :-

(i) Dual behaviour of matter,

(ii) Heisenberg uncertainty principle.

(i) Dual behaviour of matter,

(ii) Heisenberg uncertainty principle.

de Broglie (1924) proposed that matter, like radiation also exhibit dual behaviour i.e. both particle and wave like properties. de Broglie gave the relation between wavelength (`lamda`) and momentum (`p`) of a material particle as

`lamda = h/(mv) = h/p`

`m` = mass of electron , `v` = its velocity , `p` = its momentum

Note : The wave nature of electron is used in the making of electron microscope.

According to de Broglie, every object in motion has wave character. The wavelength associated with ordinaryobjects are so short that their wave properties cannot be detected.

`lamda = h/(mv) = h/p`

`m` = mass of electron , `v` = its velocity , `p` = its momentum

Note : The wave nature of electron is used in the making of electron microscope.

According to de Broglie, every object in motion has wave character. The wavelength associated with ordinaryobjects are so short that their wave properties cannot be detected.

Q 2612234139

What will be the wavelength of a ball of mass `0.1` kg moving with a velocity of `10 m s^(–1) ?`

According to de Brogile equation

`lamda = h/(mv) = (6.626xx10^(-34) Js)/((0.1kg)(10 ms^(-1)))`

` = 6.626xx10^(-34) m(J = kg m^2 s^(-2))`

Q 2622334231

The mass of an electron is `9.1xx10^(-31)kg.` If its K.E. is `3.0xx10^(-25) J` calculate its wavelength

Since `K. E. = ½ mv^2`

`v = ((2K.E.)/m)^(1/2) = ((2xx3.0xx10^(-25) kgm^2 s^(-2))/(9.1xx10^(-31) kg))^(1/2)`

` = 812 ms^(-1)`

`lamda = h/(mv)`

` = (6.626xx10^(-34) Js)/((9.1xx10^(-31))(812 ms^(-1))`

` = 8967xx10^(-10) m = 896.7nm`

Q 2652334234

Calculate the mass of a photon with wavelength `3.6 A^0`.

`lamda = 3.6A^0 = 3.6xx10^(-10)m`

Velocity of photon = velocity of light

`m = h/(lamda v)`

` = (6.626xx10^(-34) Js)/((3.6xx10^(-10)m)(3xx10^8 ms^(-1)))`

` = 6.135xx10^(-29) kg`

Defination : It states that it is impossible to determine simultaneously the exact position and exact momentum (or velocity) of an electron.

Note : - This is the result of dual behaviour of matter and radiation.

Mathematically, it can be expressed as : -

`Delta x xx Delta p_x >= h/(4 pi)`

or ` Delta x xx Delta (mv_x) >= h/(4 pi)`

or ` Delta x xx Delta v_x >= h/(4 pi m)`

where ` Delta x =` uncertainty in position

` Delta p_x =` uncertainty in momentum

` Delta v_x =` uncertainty in velocity

Note : - This is the result of dual behaviour of matter and radiation.

Mathematically, it can be expressed as : -

`Delta x xx Delta p_x >= h/(4 pi)`

or ` Delta x xx Delta (mv_x) >= h/(4 pi)`

or ` Delta x xx Delta v_x >= h/(4 pi m)`

where ` Delta x =` uncertainty in position

` Delta p_x =` uncertainty in momentum

` Delta v_x =` uncertainty in velocity

Q 2682445337

A microscope using suitable photons is employed to locate an electron in an atom within a distance of `0.1 A^0`. What is the uncertainty involved in the measurement of its velocity?

`Deltax . Deltap = h/(4pi ) ` or `Deltax . mDeltav = h/(4pi)`

`Deltav = h/(4pi Deltaxm)`

`Deltav = (6.626xx10^(-34) Js)/(4xx3.14xx0.1xx10^(-10)mxx 9.11xx10^(-31)kg)`

` = 0.579xx10^7 ms^(-1) (1J = 1kgm^2 s^(-2))`

` = 5.79xx10^6 ms^(-1)`

(i) For getting accurate results, the instrument used should have smaller units then the actual units of the thing to be measured.

(ii) For electron, the dimension of electron is very small i.e. it is considered as a point charge, so if we radiate electron with a light of having wavelength smaller than the dimension of electron, we can accurately measure the position of electron. The light having smaller wavelength will have higher energy. But as the energy of the electrons would change after collision, so we would be able to know very little about the velocity of the electron.

(ii) For electron, the dimension of electron is very small i.e. it is considered as a point charge, so if we radiate electron with a light of having wavelength smaller than the dimension of electron, we can accurately measure the position of electron. The light having smaller wavelength will have higher energy. But as the energy of the electrons would change after collision, so we would be able to know very little about the velocity of the electron.

Important Conclusion -

(i) It rules out existence of definite paths or trajectories of electrons and other similar particles.

(ii) The trajectories of an object is measured by its location and velocity.

If position of a body is known at some instant and also we know its velocity and forces acting on it at the same time we can interpret where the body would be after sometime. Since it is not possible for a sub-atomic particle to measure both position and velocity precisely and accurately, we cannot tell about its trajectory.

(iii) The principle is not applicable to macroscopic objects.

(i) It rules out existence of definite paths or trajectories of electrons and other similar particles.

(ii) The trajectories of an object is measured by its location and velocity.

If position of a body is known at some instant and also we know its velocity and forces acting on it at the same time we can interpret where the body would be after sometime. Since it is not possible for a sub-atomic particle to measure both position and velocity precisely and accurately, we cannot tell about its trajectory.

(iii) The principle is not applicable to macroscopic objects.

Q 2612545430

A golf ball has a mass of 40g, and a speed of 45 m/s. If the speed can be measured within accuracy of 2%, calculate the uncertainty in the position?

The uncertainty in the speed is 2%, i.e.,

`45 xx 2/100 = 0.9ms^(-1)`

Using the equation

`Deltax = h/(4pi m Deltav)`

` = (6.626xx10^(-34) Js)/(4xx3.14xx40gxx10^(-3) kg g^(-1) (0.9 ms^(-1)))`

` = 1.46xx10^(-33) m`

This is nearly `~ 10^(18)` times smaller than the diameter of a typical atomic nucleus. For large particles, the uncertainty principle sets no meaningful limit to the precision of measurements.