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postnew [5]
3 years ago
12

If any 1 studying 11th can u plz send me the chemistry notes for chapter 2 STRUCTURE OF AN ATOM lesson

Chemistry
2 answers:
maxonik [38]3 years ago
8 0

Answer:

MARK AS BRAINLIST IF IT IS USEFUL

Explanation:

Discovery of Electron—Discharge Tube Experiment

In 1879, William Crooks studied the conduction of electricity through gases at low pressure. He performed the experiment in a discharge tube which is a cylindrical hard glass tube about 60 cm in length. It is sealed at both the ends and fitted with two metal electrodes as shown in Fig. 2.1.

The electrical discharge through the gases could be observed only at very low pressures and at very high voltages.

The pressure of different gases could be adjusted by evacuation. When sufficiently high voltage is applied across the electrodes, current starts flowing through a stream of particles moving in the tube from the negative electrode (cathode) to the positive electrode (anode). These were called cathode rays or cathode ray particles.

• Properties of Cathode Rays

(i) Cathode rays travel in straight line.

(ii) Cathode rays start from cathode and move towards the anode.

(iii) These rays themselves are not visible but their behaviour can be observed with the help of certain kind of materials (fluorescent or phosphorescent) which glow when hit by them.

(iv) Cathode rays consist of negatively charged particles. When electric field is applied on the cathode rays with the help of a pair of metal plates, these are found to be deflected towards the positive plate indicating the presence of negative charge.

(v) The characteristics of cathode rays do not depend upon the material of electrodes and the nature of gas present in the cathode ray’tube.

• Determination of Charge/Mass (elm) Ratio for Electrons

J. J. Thomson for the first time experimentally determined charge/mass ratio called elm ratio for the electrons. For this, he subjected the beam of electrons released in the discharge tube as cathode rays to influence the electric and magnetic fields. These were acting perpendicular to one another as well as to the path followed by electrons.

According to Thomson, the amount of deviation of the particles from their path in presence of electrical and magnetic field depends upon following factors:

(i) Greater the magnitude of the charge on the particle, greater is the interaction with the electric or magnetic field and thus greater is the deflection.

(ii) The mass of the particle — lighter the particle, greater the deflection.

(iii) The deflection of electrons from their original path increases with the increase in the voltage across the electrodes or strength of the magnetic field.

By carrying out accurate measurements on the amount of deflections observed by the electrons on the electric field strength or magnetic field strength, Thomson was able to determine the value of

e/me = 1.758820 x 1011 C kg-1 where me = Mass of the electron in kg

e = magnitude of charge on the electron in coulomb (C).

• Charge on the Electron

R.A. Millikan devised a method known as oil drop experiment to determine the charge on the electrons.

• Discovery of Proton—Anode Rays

In 1886, Goldstein modified the discharge tube by using a perforated cathode. On reducing the pressure, he observed a new type of luminous rays passing through the holes or perforations of the cathode and moving in a direction opposite to the cathode rays. These rays were named as positive rays or anode rays or as canal rays. Anode rays are not emitted from the anode but from a space between anode and cathode.

• Properties of Anode Rays

(i) The value of positive charge (e) on the particles constituting anode rays depends upon the nature of the gas in the discharge tube.

(ii) The charge to mass ratio of the particles is found to depend on the gas from which these originate.

(iii) Some of the positively charged particles carry a multiple of the fundamental unit of electrical charge.

(iv) The behaviour of these particles in the magnetic or electric field is opposite to that observed for electron or cathode rays.

• Proton

The smallest and lightest positive ion was obtained from hydrogen and was called proton. Mass of proton = 1.676 x 10-27 kg

Charge on a proton = (+) 1.602 x 10-19 C

• Neutron

It is a neutral particle. It was discovered by Chadwick (1932).

By the bombardment of thin sheets of beryllium with fast moving a-particles he observed • that highly penetrating rays consist of neutral particles which were named neutrons.

• Thomson Model of Atom

timofeeve [1]3 years ago
3 0

Answer:

Atom is the smallest particle of matter that can take part in a chemical reaction.

Atom is made of electron, proton and neutrons.

Atom is not capable of independent existence.

Two or more atoms combine together to form molecules.

Electron:

Negatively charged particle discovered on the basis of ‘cathode ray discharge tube’ experiments.

Conclusion from ‘cathode ray discharge tube’ experiment:

Cathode rays start from cathode and move toward anode.

These rays are not visible but their behaviour can be observed with fluorescent or phosphorus sent material.

In the absence of electric or magnetic field these travels in strait lines.

In the presence of electric or magnetic field the behaviour of cathode rays is similar to negatively charged particles which suggest that these rays contain negatively charge particles called electron.

Proton:

Positively charged particle discovered on the basis of anode ray experiment.

Some of the characteristics of anode rays, also called canal rays, are:

These travel in straight line and posses mass many times the mass of an electron.

These are not originated from anode.

These are deflected by electric and magnetic field.

Unlike cathode rays, the positively charged particles depend upon the nature of the gas from which these originate.

Neutron:

Neutral particles discovered by bombarding a thin sheet of beryllium by α- particles.

Conclusion from α- particles scattering experiment:

Most of the α-particles passed through foil undeflected, indicating most of the space in atom is empty.

Some of the α-particles are deflected to certain angles, which means that there is positively mass present in atom.

Only some of the α-particles suffered large deflections, which means that the positively charged mass must be occupying very small space.

Strong deflections or even bouncing back of α-particles from metal foil indicate the direct collision with positively charged mass in atom.

Comparison between the subatomic particles of an atom

subatomic particles comparison

Thomson model of atom:

atom is considered asa uniform positively charged sphere with radius about 10-10 m, in which electrons are is uniformly distributed.

Electrons are embedded in such a manner to give most stable electrostatic arrangement.

Mass of atom is assumed to be uniformly distributed in atom.

Also known as plum pudding raisin pudding orwatermelon model.

Rutherford’s Nuclear Model of Atom:

Based upon α-particles scattering experiment.

Most part of the atom is empty.

Atom possesses a highly dense, positively charged centre, called nucleus of the order 10-13 cm.

Entire mass of the atom is concentrated inside the nucleus.

Electrons revolve around the nucleus in circular orbits.

Electrons and the nucleus are held together by electrostatic forces of attraction.

Drawbacks of Rutherford’s Model:

It doesn’t explain the stability of atom.

It doesn’t say anything about the electronic distribution of electrons around nucleus.

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TEA [102]
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3 years ago
Something that is made out of natural resources for human use is called
ANEK [815]

Answer:

Man-made resources

Explanation:

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6 0
3 years ago
How many atoms are in 35.1 g of sodium chloride (NaCl) ?
hoa [83]

Answer:

I believe that there are 70.2

Explanation:

There are 2 atoms in sodium chloride so I x 35.1 by 2 and got 70.2

8 0
3 years ago
Can you help me plz I think it’s b or c
Anettt [7]

i think its C pls dont get mad if it is wrong

4 0
3 years ago
Read 2 more answers
Help fast !
jok3333 [9.3K]

Answer:

1 mole represents 6.023×1023 particles.

1 mole of iodine atom= 6.023×1023

Given 127.0g of iodine.

no. of iodine atom = 1 mole of iodine

1mole of magnesium = 24g of Mg = 6.023×1023no.of Mg

Given 48g of Mg = 2×6.023×1023

no. of Mg = 2 moles of Mg

1 mole of chlorine atom= 6.023× 1023

no. of chlorine atom = 35.5g of chlorine atom

Given 71g of chlorine atom=2× 6.023× 1023

no. of chlorine atom = 6.023×1023

2 moles of chlorine atom.

Given that 4g of hydrogen atom.

will be equal to 4 × 6.023 × 1023

no. of atoms of hydrogen= 4 moles of hydrogen atom.

7 0
3 years ago
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