Chemistry Rutherford experiment

Question 5.1: How did Rutherford’s model of an atom first of al proved existence of nucleus of the atom?

Answer: In 1911, Rutherford performed an experiment in which he studied the scattering of alpha particles. When a beam of alpha particles was bombarded on 0.00004cm thick gold foil. It was observed that most of particles passed through foil undeflected and struck the ZnS coated screen used as detector. Some particles were deflected at fairly large angles and few were deflected backward. Rutherford proposed that renounced particles must have collided with central heavy portion of atom which he called nucleus.

Figure: Rutherford experiment

Question 5.2: What are the defects of Rutherford’s atomic model?

Answer: Rutherford’s planet like model was defective and unsatisfactory because: i. It was based on the laws of motions and gravitation. These laws are not for charged particles ii. The moving electron must be accelerated towards nucleus and radius of electron should become smaller and smaller and ultimately fall into nucleus. Thus, atom should collapse but actually it never happens. iii. The atom should give continuous spectrum, but they give line spectra.

Question 5.3: What are main points of Plank’s quantum theory? How can the energy of the photon be calculated from the frequency, wave length or wave number? OR what are relationships between energy of photon and frequency, wave length and wave number?

Answer: The main points of Planck’s quantum theory are:

i. Energy is not emitted or absorbed continuously. Rather, it is emitted or absorbed in a discontinuous manner in the form of wave packets or quantum. In case of light these wave packets are called photons.

ii. Amount of energy (E) associated with quantum of radiation is proportional to the frequency (V) of the radiation. Frequency is number of waves passing through a point per second.

Where ‘h’ is a plank’s constant and its values is 6.626 x10-34 J.s

iii. The frequency ‘ ‘ is related to the wave length of the photon as

v =c/λ Greater the wave length, smaller the frequency of photon.

Wave length is the distance between the two adjacent crests or troughs. By putting the value of v in equations (1) we get

E=hc/ λ ………… (2) Wave number is reciprocal of wave length and is number of waves per unit length. (v ) ̅=1/ λ or λ=1/ (v ) ̅ Putting the value of λ in equation 3,

we get

E =hc(v ) ̅ ……………… (3) From equations (1), (2) and (3), show that how energy is related to the frequency, wave length, and wave number.

Question 5.4: Drive an expression showing that the energy of photon is related to its frequency, wave length and wave number.

Answer: According to one of postulates of Planck’s quantum theory

E ∝V or E =hv ………. (1) Where ‘h’ is a plank’s constant and its values is 6.626 x10-34 J.s The frequency ‘v ‘ is related to the wave length of the photon as

v =c/λ By putting the value of v in equations (1) we get

E=hc/ λ ………… (2) Wave number is reciprocal of wave length.ie

(v ) ̅=1/ λ or λ=1/ (v ) ̅ Putting the value of λ in equation 3, we get

E =hc(v ) ̅ ……………… (3) From equations (1), (2) and (3), show that how energy is related to the frequency, wave length, and wave number.

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