Q51.Consider an imaginary ion 48X322 - . The nucleus contains 'a' % more neutrons than the number of electrons in the ion. The value of 'a' is
What This Question Tests
This question tests the calculation of the number of protons, neutrons, and electrons in an ion and then using these values to determine a percentage difference.
Concepts Tested
Formulas Used
Number of neutrons = Mass number - Atomic number
Number of electrons = Atomic number - charge
π NCERT Sections This Tests
75.4 β And 24.6 Per Cent, Respectively. Thus, The Average Mass Of A Chlorine
Physics Class 12 Β· Chapter 75
75.4 and 24.6 per cent, respectively. Thus, the average mass of a chlorine atom is obtained by the weighted average of the masses of the two isotopes, which works out to be 75.4 Γ 34.98 + 24.6 Γ 36.98 = 100 = 35.47 u which agrees with the atomic mass of chlorine. Even the lightest element, hydrogen has three isotopes having masses 1.0078 u, 2.0141 u, and 3.0160 u. The nucleus of the lightest atom of hydrogen, which has a relative abundance of 99.985%, is called the proton. The mass of a proton is β27 m p = 1.00727 u = 1.67262 Γ 10 kg (13.2) This is equal to the mass of the hydrogen atom (= 1.00783u), minus the mass of a single electron (me = 0.00055 u). The other two isotopes of hydrogen are called deuterium and tritium. Tritium nuclei, being unstable, do not occur naturally and are produced artificially in laboratories. The positive charge in the nucleus is that of the protons. A proton carries one unit of fundamental charge and is stable. It was earlier thought that the nucleus may contain electrons, but this was ruled out later using arguments based on quantum theory. All the electrons of an atom are outside the nucleus. We know that the number of these electrons outside 307the nucleus of the atom is Z, the atomic number. The total charge of the Reprint 2025-26 Physics atomic electrons is thus (βZe), and since the atom is neutral, the charge of the nucleus is (+Ze). The number of protons in the nucleus of the atom is, therefore, exactly Z, the atomic number. Discovery of Neutron Since the nuclei of deuterium and tritium are isotopes of hydrogen, they must contain only one proton each. But the masses of the nuclei of hydrogen, deuterium and tritium are in the ratio of 1:2:3. Therefore, the nuclei of deuterium and tritium must contain, in addition to a proton, some neutral matter. The amount of neutral matter present in the nuclei of these isotopes, expressed in units of mass of a proton, is approximately equal to one and two, respectively. This fact indicates that the nuclei of atoms contain, in addition to protons, neutral matter in multiples of a basic unit. This hypothesis was verified in 1932 by James Chadwick who observed emission of neutral radiation when beryllium nuclei were bombarded with alpha-particles (a-particles are helium nuclei, to be discussed in a later section). It was found that this neutral radiation could knock out protons from light nuclei such as those of helium, carbon and nitrogen. The only neutral radiation known at that time was photons (electromagnetic radiation). Application of the principles of conservation of energy and momentum showed that if the neutral radiation consisted of photons, the energy of photons would have to be much higher than is available from the bombardment of beryllium nuclei with a-particles. The clue to this puzzle, which Chadwick satisfactorily solved, was to assume that the neutral radiation consists of a new type of neutral particles called neutrons. From conservation of energy and momentum, he was able to determine the mass of new particle βas very nearly the same as mass of protonβ. The mass of a neutron is now known to a high degree of accuracy. It is m n = 1.00866 u = 1.6749Γ10β27 kg (13.3) Chadwick was awarded the 1935 Nobel Prize in Physics for his discovery of the neutron. A free neutron, unlike a free proton, is unstable. It decays into a proton, an electron and a antineutrino (another elementary particle), and has a mean life of about 1000s. It is, however, stable inside the nucleus. The composition of a nucleus can now be described using the following terms and symbols: Z - atomic number = number of protons [13.4(a)] N - neutron number = number of neutrons [13.4(b)] A - mass number = Z + N = total number of protons and neutrons [13.4(c)] One also uses the term nucleon for a proton or a neutron. Thus the number of nucleons in an atom is its mass number A. Nuclear species or nuclides are shown by the notation ZA X where X is the chemical symbol of the species. For example, the nucleus of gold is denoted by 19779 Au . It contains 197 nucleons, of which 79 are protons 308 and the rest118 are neutrons. Reprint 2025-26 Nuclei The composition of isotopes of an element can now be readily explained. The nuclei of isotopes of a given element contain the same number of protons, but differ from each other in their number of neutrons. Deuterium, 12 H, which is an isotope of hydrogen, contains one proton and one neutron. Its other isotope tritium, 13 H, contains one proton and two neutrons. The element gold has 32 isotopes, ranging from A =173 to A = 204. We have already mentioned that chemical properties of elements depend on their electronic structure. As the atoms of isotopes have identical electronic structure they have identical chemical behaviour and are placed in the same location in the periodic table. All nuclides with same mass number A are called isobars. For example, the nuclides 13 H and 32He are isobars. Nuclides with same neutron number N but different atomic number Z, for example 19880 Hg and 19779 Au , are called isotones.
13.2 β Obtain The Binding Energy Of The Nuclei 5626Fe And 20983 Bi In Units Of
Physics Class 12 Β· Chapter 13
13.2 Obtain the binding energy of the nuclei 5626Fe and 20983 Bi in units of MeV from the following data: m ( 5626Fe ) = 55.934939 u m ( 20983 Bi ) = 208.980388 u
12.5 β A Hydrogen Atom Initially In The Ground Level Absorbs A Photon,
Physics Class 12 Β· Chapter 12
12.5 A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.
π Question Details
- Chapter
- Atomic Structure
- Topic
- Atomic number, mass number, isotopes
- Year
- 2022
- Shift
- 26 Jul Shift 2
- Q Number
- Q51
- Type
- Numerical
- NCERT Ref
- Class 11 Chemistry Ch 2: Atomic Structure
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