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Zorluk: OrtaRadioactivity, Nuclear Reactions, and Half-Life

Complete the statements regarding the artificial nuclear transmutation of aluminium-27 by filling in the blanks with the correct numerical values.

Cevap:When aluminium-27 (1327Al^{27}_{13}\text{Al}) is bombarded with an alpha particle (24α^{4}_{2}\alpha), it produces phosphorus-30 (1530P^{30}_{15}\text{P}) and a neutron (ZAn^{A}_{Z}\text{n}). According to the law of conservation of mass number and atomic number, the mass number AA of the emitted neutron is 【1】 and its atomic number ZZ is 【0】.

Cevap

The mass number A is 1 and the atomic number Z is 0.
In any balanced nuclear reaction, the sum of the superscripts (mass numbers) and the sum of the subscripts (atomic numbers) must be equal on both sides of the equation. For the reactants (1327Al+24α^{27}_{13}\text{Al} + ^{4}_{2}\alpha), the total mass number is 27+4=3127 + 4 = 31 and the total atomic number is 13+2=1513 + 2 = 15. For the products (1530P+ZAn^{30}_{15}\text{P} + ^{A}_{Z}\text{n}), setting 30+A=3130 + A = 31 yields A=1A = 1, and setting 15+Z=1515 + Z = 15 yields Z=0Z = 0. Thus, the emitted particle is a neutron (01n^{1}_{0}\text{n}).

Adım Adım Çözüm

1
Balance the total mass numbers on both sides of the nuclear equation.
Total mass number on reactants side = 27 + 4 = 31. Mass number of phosphorus-30 = 30. Therefore, mass number of the neutron A = 31 - 30 = 1.
The total sum of mass numbers must remain equal before and after a nuclear reaction.
2
Balance the total atomic numbers (nuclear charge) on both sides of the nuclear equation.
Total atomic number on reactants side = 13 + 2 = 15. Atomic number of phosphorus-30 = 15. Therefore, atomic number of the neutron Z = 15 - 15 = 0.
The total sum of atomic numbers (positive charges) must be conserved in nuclear reactions.

Anahtar Kavram

Conservation of mass number and atomic number in nuclear transmutation equations.
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