
A nucleus of an element \[_{84}{X^{202}}\] emits an α-particle first, a β-particle next, and then a gamma photon. The final nucleus formed has an atomic number
A. 200
B. 199
C. 83
D. 198
Answer
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Hint: The nucleus is a collection of particles referred to as protons, which are positively charged, and neutrons, which are electrically neutral. Neutrons and protons are in turn made up of particles known as quarks. When a radioactive element emits α-particles then its mass number is reduced by 4 and atomic number is reduced by 2 and when it emits β-particle then its atomic number is increased by one but mass number remains the same.
Complete step by step solution:
The term nucleon is used for either a neutron or a proton. The simplest nucleus is that of hydrogen, which is just a single proton, whereas the largest nucleus studied has approximately 300 nucleons. A nucleus is identified as in the example below by its atomic number Z (i.e., the number of protons), the mass number A and the neutron number N, where A = Z + N. The convention for specifying nuclei is by atomic number Z, and mass number A, and its chemical symbol. The neutron number is calculated by N = A - Z.
Given data,
The mass number of nucleus X = 202 and atomic number of nucleus X = 84. When a radioactive nucleus \[_{84}{X^{202}}\] emits an α-particle \[({}_4^2He)\] the mass number reduces by 4 units and the atomic number reduces by 2 units.
Finding mass number and atomic number of nucleus Y.
The mass number of nucleus Y = 202 – 4 = 198
And, the atomic number of nucleus X = 84 – 2 = 82
Write the nuclear equation
The nuclear equation is expressed as,
\[_{84}{X^{202}} \longrightarrow\, _{82}{Y^{198}}\] + \[({}_4^2He)\] + energy
Hence, the formation of nucleus Y of atomic number 82 and mass number is
\[_{84}{X^{202}} \longrightarrow\, _{82}{Y^{198}}\] + \[({}_4^2He)\]
And
\[_{82}{Y^{198}} \longrightarrow\,_{83}{Z^{198}}\] + \[_{ - 1}{\beta ^0}\]
Therefore the correct answer is option C.
Note: In addition to its mass number and atomic number, a nucleus is also defined by its shape, size, angular momentum, binding energy, and (if it is unstable) half-life. One of the best methods to specify the size of a nucleus is to disperse high-energy electrons from it. The angular distribution of the dispersed electrons is dependent on the proton distribution. The proton distribution can be described by an average radius.
Complete step by step solution:
The term nucleon is used for either a neutron or a proton. The simplest nucleus is that of hydrogen, which is just a single proton, whereas the largest nucleus studied has approximately 300 nucleons. A nucleus is identified as in the example below by its atomic number Z (i.e., the number of protons), the mass number A and the neutron number N, where A = Z + N. The convention for specifying nuclei is by atomic number Z, and mass number A, and its chemical symbol. The neutron number is calculated by N = A - Z.
Given data,
The mass number of nucleus X = 202 and atomic number of nucleus X = 84. When a radioactive nucleus \[_{84}{X^{202}}\] emits an α-particle \[({}_4^2He)\] the mass number reduces by 4 units and the atomic number reduces by 2 units.
Finding mass number and atomic number of nucleus Y.
The mass number of nucleus Y = 202 – 4 = 198
And, the atomic number of nucleus X = 84 – 2 = 82
Write the nuclear equation
The nuclear equation is expressed as,
\[_{84}{X^{202}} \longrightarrow\, _{82}{Y^{198}}\] + \[({}_4^2He)\] + energy
Hence, the formation of nucleus Y of atomic number 82 and mass number is
\[_{84}{X^{202}} \longrightarrow\, _{82}{Y^{198}}\] + \[({}_4^2He)\]
And
\[_{82}{Y^{198}} \longrightarrow\,_{83}{Z^{198}}\] + \[_{ - 1}{\beta ^0}\]
Therefore the correct answer is option C.
Note: In addition to its mass number and atomic number, a nucleus is also defined by its shape, size, angular momentum, binding energy, and (if it is unstable) half-life. One of the best methods to specify the size of a nucleus is to disperse high-energy electrons from it. The angular distribution of the dispersed electrons is dependent on the proton distribution. The proton distribution can be described by an average radius.
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