Q: The average atomic mass of a sample of an element \$X\$ is \$16.2\ u\$. What are the percentages of isotopes \$_{ 8 }^{ 16 }{ X }\$ and \$_{ 8 }^{ 18 }{ X }\$ in the sample?

Let the percentage of \$_8^{16}X\$ be \$A\$ %. Then the percentage of \$_8^{18}X\$ be \$(100-A)\$ %.\$\therefore\$ \$16\times\dfrac{A}{100}+18\times\dfrac{100-A}{100}=16.2\$\$\implies 1800-2A=1620\$\$\implies A=90\$So the answer is 90% \$_8^{16}X\$ and 10% \$_8^{18}X\$

Q: What are isotopes? Give one example.

Isotope \$\rightarrow\$ Isotopes are atoms with the same number of protons, but differ in numbers of neutrons. Isotopes are different forms of a single element.Example - Carbon 12 and Carbon 14 are both isotopes of carbon, one with 6 neutrons and one with 8 neutrons.

Q: Calculate the number of protons, electrons, neutrons in Bromine with atomic number 35 and mass number 80.

Atomic number \$z = 35\$Mass number\$=80\$No. of proton \$p=35\$No. of neutron \$n=80-35=45\$No. of electron \$e=35\$

Q: Atomic mass of chlorine is 35.5. It has two isotopes of atomic mass 35 and 37. The percentage of heavier isotope is:

Let percentage of heavier isotope be x. Thus \$35.5\times 100=37 \times x +35(100-x)\$\$\Rightarrow x=25\$%Hence answer is option D

Q: The atomic weight of Ne is 20.2. Ne is mixture of \$\displaystyle Ne^{20}\$ and \$\displaystyle Ne^{22}\$. The relative abundance of the heavier isotope is:

Let the relative abundance % of \$Ne-20 =x_1\$So, relative abundance % of \$Ne-22 =x_2\$Now we know that atomic mass = sum of relative abundance in atomic mass of each isotope / 100%\$\displaystyle 20.2=\frac{20x_{1}+22x_{2}}{x_{1}+x_{2}}\$\$\displaystyle 20.2x_{1}+20.2x_{2}=20x_{1}+22x_{2}\$\$\displaystyle 0.2x_{1}=1.8\: \: x_{2}\$ \$\displaystyle \frac{x_{1}}x_{2}{=\frac{1.8}{0.2}}=\frac{9}{1}\$ Relative abundance\$\displaystyle x_{1}\%= \frac{9}{10}\times 100= 90\%\$ lighter isotopes\$\displaystyle x_{1}\%=\frac{1}{10}\times 100=10\%\$ = heavier isotopeHence, the correct option is \$D\$

Q: Chlorine exists in two isotopic forms, \$Cl-37\$ and \$Cl-35\$ but its atomic mass is \$35.5\$. This indicates the ratio of \$Cl-37\$ and \$Cl-35\$ is approximately:

Let relative abundance of Cl-37 be \$A_1\$% with mass \$M_1= 37\$ amu\$\therefore\$ The relative abundance of Cl-35 = \$A_2\$% with mass, \$M_2=35\$ amuwhere \$A_1+A_2=100\$Thus \$A_2\$%=\$(100-A_1)\$%Using average atomic mass, \$M_{avg}=\frac { { M }_{ 1 }{ A }_{ 1 }+{ M }_{ 2 }{ A }_{ 2 } }{ { A }_{ 1 }+{ A }_{ 2 } } \$\$M_{avg} =\frac { 37 \times { A }_{ 1 }+35 \times (100-{ A }_{ 1 }) }{ { A }_{ 1 }+(100-{ A }_{ 1 }) } =\frac{2A_1+3500}{100}\$Given that: \$M_{avg}=35.5\$\$\therefore \frac{2A_1+3500}{100}=35.5\$\$2A_1=35.5 \times 100 -3500=50\$\$A_1=\frac{50}{2}=25\$ and \$A_2=100-A_1=100-25\$ =75 Thus the ratio of isotope abundance of \$Cl-37 to Cl-35\$ is \$A_1:A_2= 25: 75 = 1: 3\$So, the correct option is \$C\$

Q: Which of the following isotopes is used for treatment of cancer?

\$Co^{60}\$Cobalt therapy or cobalt -60 therapy is the medical use of gamma rays from the radioisotope cobalt 60 to treat conditions such as cancer.

Question 1

The average atomic mass of a sample of an element $X$ is $16.2\ u$. What are the percentages of isotopes $_{ 8 }^{ 16 }{ X }$ and $_{ 8 }^{ 18 }{ X }$ in the sample?

Accepted Answer

Let the percentage of $_8^{16}X$ be $A$ %. Then the percentage of $_8^{18}X$ be $(100-A)$ %.$\therefore$ $16\times\dfrac{A}{100}+18\times\dfrac{100-A}{100}=16.2$$\implies 1800-2A=1620$$\implies A=90$So the answer is 90% $_8^{16}X$ and 10% $_8^{18}X$

Question 2

Explain with examples (i) Atomic number, (ii) Mass number, (iii) Isotopes and (iv) Isobars. Give any two uses of isotopes.

Accepted Answer

(i) Atomic number: The atomic number is the total number of protons present in the atom. For example, the atomic number of sodium is 11. It contains 11 protons and 11 electrons. (ii) Mass number: It is the sum of the number of neutrons and the number of protons. For example, the atomic number of magnesium is 12 which is equal to the number of protons, the number of neutrons of magnesium is 12. The mass number is equal to 24 (12+12).(iii) Isotopes are the elements having same atomic number but different mass number. They have same chemical properties as the number of valence electrons are same. For example, $_6C^{12}, \ _6C^{13}, \ _6C^{14}$ are the three isotopes of carbon. They have same atomic number but different mass number due to difference in the number of neutrons.(iv) Isobars: They are elements having the same mass number but different atomic number. For example. $_{18}Ar^{40},\  _{19}K^{40}$ are isobars having the same atomic mass but different atomic number.The two uses of isotopes are:Nuclear weapons and nuclear power require large quantities of isotopes of uranium.An isotope of cobalt is used in the treatment of cancer.

Question 3

What are isotopes? Give one example.

Accepted Answer

Isotope $\rightarrow$ Isotopes are atoms with the same number of protons, but differ in numbers of neutrons. Isotopes are different forms of a single element.Example - Carbon 12 and Carbon 14 are both isotopes of carbon, one with 6 neutrons and one with 8 neutrons.

Question 4

Composition of the nuclei of two atomic species $X$ and $Y$ are given as under
$X$ $Y$
Protons $=$ $6$ $6$
Neutrons $=$ $6$ $8$

Given the mass numbers of $X$ and $Y$ . What is the relation between the two species?

Accepted Answer

Atomic mass=number of protons+number of neutronsAtomic mass of X = 6+6=12Atomic mass of Y =6+8=14Both X and Y have same atomic number but different atomic mass number so they are isotopes.

Question 5

Calculate the number of protons, electrons, neutrons in Bromine with atomic number 35 and mass number 80.

Accepted Answer

Atomic number $z = 35$Mass number$=80$No. of proton $p=35$No. of neutron $n=80-35=45$No. of electron $e=35$

Question 6

Atomic mass of chlorine is 35.5. It has two isotopes of atomic mass 35 and 37. The percentage of heavier isotope is:

Accepted Answer

Let percentage of heavier isotope be x. Thus $35.5\times 100=37 \times x +35(100-x)$$\Rightarrow x=25$%Hence answer is option D

Question 7

The atomic weight of Ne is 20.2. Ne is mixture of $\displaystyle Ne^{20}$ and $\displaystyle Ne^{22}$. The relative abundance of the heavier isotope is:

Accepted Answer

Let the relative abundance % of $Ne-20 =x_1$So, &#8203;relative abundance % of $Ne-22 =x_2$Now we know that atomic mass = sum of relative abundance in atomic mass of each isotope / 100%$\displaystyle 20.2=\frac{20x_{1}+22x_{2}}{x_{1}+x_{2}}$$\displaystyle 20.2x_{1}+20.2x_{2}=20x_{1}+22x_{2}$$\displaystyle 0.2x_{1}=1.8\: \: x_{2}$ $\displaystyle \frac{x_{1}}x_{2}{=\frac{1.8}{0.2}}=\frac{9}{1}$              Relative abundance$\displaystyle x_{1}\%= \frac{9}{10}\times 100= 90\%$ lighter isotopes$\displaystyle x_{1}\%=\frac{1}{10}\times 100=10\%$ = heavier isotopeHence, the correct option is $D$

Question 8

Chlorine exists in two isotopic forms, $Cl-37$ and $Cl-35$ but its atomic mass is $35.5$. This indicates the ratio of $Cl-37$ and $Cl-35$ is approximately:

Accepted Answer

Let relative abundance of Cl-37 be $A_1$% with mass $M_1= 37$ amu$\therefore$ The relative abundance of Cl-35 = $A_2$% with mass, $M_2=35$ amuwhere $A_1+A_2=100$Thus $A_2$%=$(100-A_1)$%Using average atomic mass, $M_{avg}=\frac { { M }_{ 1 }{ A }_{ 1 }+{ M }_{ 2 }{ A }_{ 2 } }{ { A }_{ 1 }+{ A }_{ 2 } } $$M_{avg} =\frac { 37 \times { A }_{ 1 }+35 \times (100-{ A }_{ 1 }) }{ { A }_{ 1 }+(100-{ A }_{ 1 }) } =\frac{2A_1+3500}{100}$Given that: $M_{avg}=35.5$$\therefore \frac{2A_1+3500}{100}=35.5$$2A_1=35.5 \times 100 -3500=50$$A_1=\frac{50}{2}=25$ and $A_2=100-A_1=100-25$ =75 Thus the ratio of isotope abundance of $Cl-37 to Cl-35$ is $A_1:A_2= 25: 75 = 1: 3$So, the correct option is $C$

Question 9

Which of the following isotopes is used for treatment of cancer?

Accepted Answer

$Co^{60}$Cobalt therapy or cobalt -60 therapy is the medical use of gamma rays from the radioisotope cobalt 60 to treat conditions such as cancer.

Question 10

An atom of an element contains $13$ electrons. Its nucleus has $14$ neutrons. Find out its atomic number and approximate atomic mass. An isotope has atomic mass $2$ units higher. What will be the number of protons, neutrons and electrons in the isotope?

Accepted Answer

Atomic number=No. of protons=No. of electrons in ground state(unchanged atom)Atomic number=$13$=No. of protonsAtomic mass=no. of protons+no. of neutrons=$13+14=27$For isotope no. of proton=$13$(same atomic number but different mass number are isotopes)no. of electrons=$13$no. of neutrons=$14+2=16$

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Isotopes

Isotopes and Isobars-Basic

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Isobars, Isotones and Isodiaphers

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The average atomic mass of a sample of an element $X$ is $16.2 u$ . What are the percentages of isotopes $_{8}^{16} X$ and $_{8}^{18} X$ in the sample?

Explain with examples (i) Atomic number, (ii) Mass number, (iii) Isotopes and (iv) Isobars. Give any two uses of isotopes.

What are isotopes? Give one example.

Composition of the nuclei of two atomic species $X$ and $Y$ are given as under
$X$ $Y$
Protons $=$ $6$ $6$
Neutrons $=$ $6$ $8$

Given the mass numbers of $X$ and $Y$ . What is the relation between the two species?

Calculate the number of protons, electrons, neutrons in Bromine with atomic number 35 and mass number 80.

Atomic mass of chlorine is 35.5. It has two isotopes of atomic mass 35 and 37. The percentage of heavier isotope is:

The atomic weight of Ne is 20.2. Ne is mixture of $N e^{20}$ and $N e^{22}$ . The relative abundance of the heavier isotope is:

Chlorine exists in two isotopic forms, $C l - 37$ and $C l - 35$ but its atomic mass is $35.5$ . This indicates the ratio of $C l - 37$ and $C l - 35$ is approximately:

Which of the following isotopes is used for treatment of cancer?

An atom of an element contains $13$ electrons. Its nucleus has $14$ neutrons. Find out its atomic number and approximate atomic mass. An isotope has atomic mass $2$ units higher. What will be the number of protons, neutrons and electrons in the isotope?

Revise with Concepts

Basics of Isotopes, Isobars, Isotones and Isodiaphers

Isotopes

Isotopes and Isobars-Basic

Quick Summary With Stories

Isobars, Isotones and Isodiaphers

Isotopes

Radioactive Isotopes and Its Uses

The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes 816​X and 818​X in the sample?

Explain with examples (i) Atomic number, (ii) Mass number, (iii) Isotopes and (iv) Isobars. Give any two uses of isotopes.

What are isotopes? Give one example.

Composition of the nuclei of two atomic species X and Y are given as under X Y Protons = 6 6 Neutrons= 6 8 Given the mass numbers of X and Y. What is the relation between the two species?

Calculate the number of protons, electrons, neutrons in Bromine with atomic number 35 and mass number 80.

Atomic mass of chlorine is 35.5. It has two isotopes of atomic mass 35 and 37. The percentage of heavier isotope is:

The atomic weight of Ne is 20.2. Ne is mixture of Ne20 and Ne22. The relative abundance of the heavier isotope is:

Chlorine exists in two isotopic forms, Cl−37 and Cl−35 but its atomic mass is 35.5. This indicates the ratio of Cl−37 and Cl−35 is approximately:

Which of the following isotopes is used for treatment of cancer?

An atom of an element contains 13 electrons. Its nucleus has 14 neutrons. Find out its atomic number and approximate atomic mass. An isotope has atomic mass 2 units higher. What will be the number of protons, neutrons and electrons in the isotope?

The average atomic mass of a sample of an element $X$ is $16.2 u$ . What are the percentages of isotopes $_{8}^{16} X$ and $_{8}^{18} X$ in the sample?

Composition of the nuclei of two atomic species $X$ and $Y$ are given as under
$X$ $Y$
Protons $=$ $6$ $6$
Neutrons $=$ $6$ $8$

Given the mass numbers of $X$ and $Y$ . What is the relation between the two species?

The atomic weight of Ne is 20.2. Ne is mixture of $N e^{20}$ and $N e^{22}$ . The relative abundance of the heavier isotope is:

Chlorine exists in two isotopic forms, $C l - 37$ and $C l - 35$ but its atomic mass is $35.5$ . This indicates the ratio of $C l - 37$ and $C l - 35$ is approximately:

An atom of an element contains $13$ electrons. Its nucleus has $14$ neutrons. Find out its atomic number and approximate atomic mass. An isotope has atomic mass $2$ units higher. What will be the number of protons, neutrons and electrons in the isotope?