Chemistry Class 11 NCERT Solutions: Chapter 2 Structure of Atom Part 12

Q: 39 In Millikan’s experiment, static electric charge on the oil drops have been obtained by shining X-rays. If the static electric charge on the oil drop is, Equation calculate the number of electrons present on it.

Image of Millikan Oil Drop Experiment for Structure in Atom

Image of Millikan Oil Drop Experiment

Image of Millikan Oil Drop Experiment for Structure in Atom

Answer:

Charge on the oil drop Equation

Charge on one electron Equation

Equation

Equation

Equation

Equation

Q: 40. In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the Equation -particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?

Image of Rutherford's Gold Foil Experiment for Structure of Atom

Image of Rutherford's Gold Foil Experiment

Image of Rutherford's Gold Foil Experiment for Structure of Atom

Answer:

Heavy atoms have nucleus carrying large amount of positive charge. Therefore, some Equation - particles will easily deflected back. Also a number of Equation - particles are deflected through small angles because of large positive charge.

A thin foil of lighter atoms will not give the same results as given with the foil of heavier atoms.

Lighter atoms would be able to carry very little positive charge. Hence, they will not cause enough deflection of Equation -particles (positively charged).

Q: 41. Symbols Equation can be written, whereas symbols Equation are not acceptable. Answer briefly.

Answer:

The general convention of representing an element along with its atomic mass (A) and atomic number (Z) is Equation

Equation be written because the atomic number of an element is constant, but the atomic mass of an element depends upon the relative abundance of its isotopes. Hence, it is necessary to mention the atomic mass of an element.

Q: 42. An element with mass number Equation contains Equation more neutrons as compared to protons. Assign the atomic symbol.

Answer:

Let the number of protons in the element be X.

Equation

Equation

= x Equation x

Equation x

According to the question,

Mass number of the element Equation

Equation

Equation

Equation

Equation

Equation

Equation

Hence, the number of protons in the element i.e., x is 35.

Since the atomic number of an atom is defined as the number of protons present in its nucleus, the atomic number of the given element is 35.

Equation

Q: 43. An ion with mass number Equation possesses one unit of negative charge. If the ion contains Equation more neutrons than the electrons, find the symbol of the ion.

Let the number of electrons in the ion carrying a negative charge be x. Then, Number of neutrons present

Equation

Equation

Equation

Number of electrons in the neutral atom Equation

(When an ion carries a negative charge, it carries an extra electron)

∴ Number of protons in the neutral atom Equation

Given,

Mass number of the ion Equation

Equation

Equation

Equation

Therefore no of protons = atomic no Equation

∴The symbol of the ion is Equation

Q: 44. An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.

Answer:

Let the number of electrons present in ion Equation

Equation

Since the ion is tripositive,

Equation

Equation

Given,

Mass number of the ion = 56

Equation

Equation

Equation

Equation

Equation

Equation

Q: 45. Arrange the following type of radiations in increasing order of frequency:

(a) Radiation from microwave oven

(b) Amber light from traffic signal

(c) Radiation from FM radio

(d) Cosmic rays from outer space and

(e) X-rays

Answer:

The increasing order of frequency is as follows:

Radiation from FM radio <amber light < radiation from microwave oven < X-rays < cosmic rays

The increasing order of wavelength is as follows:

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

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