363839
Statement A : Neutrons penetrate matter more readily as compared to protons. Statement B : Neutrons are slightly more massive than protons.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
As neutrons have no charge so their penetration power is more than protons and electrons. So option (3) is correct.
PHXII13:NUCLEI
363840
A nucleus represented by the symbol \({ }_{Z}^{A} X\) has
1 \(Z\) neutrons and \(A-Z\) protons
2 \(Z\) protons and \(A-Z\) neutrons
3 \(Z\) protons and \(A\) neutrons
4 \(A\) protons and \(Z-A\) neutrons
Explanation:
The nucleus, \({ }_{Z} X^{A}\) has number of protons \(=Z\) and number of neutrons \(=A-Z\), where \(A\) is the total number of protons and neutrons, \(i.\,e.\,\,A = Z + N\)
PHXII13:NUCLEI
363841
If the radius of a nucleus of mass number 3 is \(R\), then the radius of a nucleus of mass number 81 is
363842
The radius of the nucleus is proportional to, (if \(A\) is the atomic mass number)
1 \(A\)
2 \({A^3}\)
3 \({A^{1/3}}\)
4 \({A^{2/3}}\)
Explanation:
\(R = {R_0}{A^{1/3}}\)
PHXII13:NUCLEI
363843
Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio of radii of nucleus to that of Helium nucleus is \({(14)^{1/3}}\). The atomic number of nucleus will be
1 25
2 26
3 56
4 30
Explanation:
The radius of a nucleus is \(R=R_{0} A^{1 / 3}\) where \(R_{0}\) is a constant and \(A\) be its mass number. \(\therefore \quad \frac{R}{{{R_{He}}}} = {\left( {\frac{A}{4}} \right)^{1/3}}\) As \(\frac{R}{{{R_{He}}}} = {(14)^{1/3}}\) (given) \(\therefore \quad(14)^{1 / 3}=\left(\dfrac{A}{4}\right)^{1 / 3} ; 14=\dfrac{A}{4}\) or \(A=56\) The atomic number of nucleus is \(Z=A-N=56-30=26\)
363839
Statement A : Neutrons penetrate matter more readily as compared to protons. Statement B : Neutrons are slightly more massive than protons.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
As neutrons have no charge so their penetration power is more than protons and electrons. So option (3) is correct.
PHXII13:NUCLEI
363840
A nucleus represented by the symbol \({ }_{Z}^{A} X\) has
1 \(Z\) neutrons and \(A-Z\) protons
2 \(Z\) protons and \(A-Z\) neutrons
3 \(Z\) protons and \(A\) neutrons
4 \(A\) protons and \(Z-A\) neutrons
Explanation:
The nucleus, \({ }_{Z} X^{A}\) has number of protons \(=Z\) and number of neutrons \(=A-Z\), where \(A\) is the total number of protons and neutrons, \(i.\,e.\,\,A = Z + N\)
PHXII13:NUCLEI
363841
If the radius of a nucleus of mass number 3 is \(R\), then the radius of a nucleus of mass number 81 is
363842
The radius of the nucleus is proportional to, (if \(A\) is the atomic mass number)
1 \(A\)
2 \({A^3}\)
3 \({A^{1/3}}\)
4 \({A^{2/3}}\)
Explanation:
\(R = {R_0}{A^{1/3}}\)
PHXII13:NUCLEI
363843
Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio of radii of nucleus to that of Helium nucleus is \({(14)^{1/3}}\). The atomic number of nucleus will be
1 25
2 26
3 56
4 30
Explanation:
The radius of a nucleus is \(R=R_{0} A^{1 / 3}\) where \(R_{0}\) is a constant and \(A\) be its mass number. \(\therefore \quad \frac{R}{{{R_{He}}}} = {\left( {\frac{A}{4}} \right)^{1/3}}\) As \(\frac{R}{{{R_{He}}}} = {(14)^{1/3}}\) (given) \(\therefore \quad(14)^{1 / 3}=\left(\dfrac{A}{4}\right)^{1 / 3} ; 14=\dfrac{A}{4}\) or \(A=56\) The atomic number of nucleus is \(Z=A-N=56-30=26\)
363839
Statement A : Neutrons penetrate matter more readily as compared to protons. Statement B : Neutrons are slightly more massive than protons.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
As neutrons have no charge so their penetration power is more than protons and electrons. So option (3) is correct.
PHXII13:NUCLEI
363840
A nucleus represented by the symbol \({ }_{Z}^{A} X\) has
1 \(Z\) neutrons and \(A-Z\) protons
2 \(Z\) protons and \(A-Z\) neutrons
3 \(Z\) protons and \(A\) neutrons
4 \(A\) protons and \(Z-A\) neutrons
Explanation:
The nucleus, \({ }_{Z} X^{A}\) has number of protons \(=Z\) and number of neutrons \(=A-Z\), where \(A\) is the total number of protons and neutrons, \(i.\,e.\,\,A = Z + N\)
PHXII13:NUCLEI
363841
If the radius of a nucleus of mass number 3 is \(R\), then the radius of a nucleus of mass number 81 is
363842
The radius of the nucleus is proportional to, (if \(A\) is the atomic mass number)
1 \(A\)
2 \({A^3}\)
3 \({A^{1/3}}\)
4 \({A^{2/3}}\)
Explanation:
\(R = {R_0}{A^{1/3}}\)
PHXII13:NUCLEI
363843
Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio of radii of nucleus to that of Helium nucleus is \({(14)^{1/3}}\). The atomic number of nucleus will be
1 25
2 26
3 56
4 30
Explanation:
The radius of a nucleus is \(R=R_{0} A^{1 / 3}\) where \(R_{0}\) is a constant and \(A\) be its mass number. \(\therefore \quad \frac{R}{{{R_{He}}}} = {\left( {\frac{A}{4}} \right)^{1/3}}\) As \(\frac{R}{{{R_{He}}}} = {(14)^{1/3}}\) (given) \(\therefore \quad(14)^{1 / 3}=\left(\dfrac{A}{4}\right)^{1 / 3} ; 14=\dfrac{A}{4}\) or \(A=56\) The atomic number of nucleus is \(Z=A-N=56-30=26\)
363839
Statement A : Neutrons penetrate matter more readily as compared to protons. Statement B : Neutrons are slightly more massive than protons.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
As neutrons have no charge so their penetration power is more than protons and electrons. So option (3) is correct.
PHXII13:NUCLEI
363840
A nucleus represented by the symbol \({ }_{Z}^{A} X\) has
1 \(Z\) neutrons and \(A-Z\) protons
2 \(Z\) protons and \(A-Z\) neutrons
3 \(Z\) protons and \(A\) neutrons
4 \(A\) protons and \(Z-A\) neutrons
Explanation:
The nucleus, \({ }_{Z} X^{A}\) has number of protons \(=Z\) and number of neutrons \(=A-Z\), where \(A\) is the total number of protons and neutrons, \(i.\,e.\,\,A = Z + N\)
PHXII13:NUCLEI
363841
If the radius of a nucleus of mass number 3 is \(R\), then the radius of a nucleus of mass number 81 is
363842
The radius of the nucleus is proportional to, (if \(A\) is the atomic mass number)
1 \(A\)
2 \({A^3}\)
3 \({A^{1/3}}\)
4 \({A^{2/3}}\)
Explanation:
\(R = {R_0}{A^{1/3}}\)
PHXII13:NUCLEI
363843
Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio of radii of nucleus to that of Helium nucleus is \({(14)^{1/3}}\). The atomic number of nucleus will be
1 25
2 26
3 56
4 30
Explanation:
The radius of a nucleus is \(R=R_{0} A^{1 / 3}\) where \(R_{0}\) is a constant and \(A\) be its mass number. \(\therefore \quad \frac{R}{{{R_{He}}}} = {\left( {\frac{A}{4}} \right)^{1/3}}\) As \(\frac{R}{{{R_{He}}}} = {(14)^{1/3}}\) (given) \(\therefore \quad(14)^{1 / 3}=\left(\dfrac{A}{4}\right)^{1 / 3} ; 14=\dfrac{A}{4}\) or \(A=56\) The atomic number of nucleus is \(Z=A-N=56-30=26\)
363839
Statement A : Neutrons penetrate matter more readily as compared to protons. Statement B : Neutrons are slightly more massive than protons.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
As neutrons have no charge so their penetration power is more than protons and electrons. So option (3) is correct.
PHXII13:NUCLEI
363840
A nucleus represented by the symbol \({ }_{Z}^{A} X\) has
1 \(Z\) neutrons and \(A-Z\) protons
2 \(Z\) protons and \(A-Z\) neutrons
3 \(Z\) protons and \(A\) neutrons
4 \(A\) protons and \(Z-A\) neutrons
Explanation:
The nucleus, \({ }_{Z} X^{A}\) has number of protons \(=Z\) and number of neutrons \(=A-Z\), where \(A\) is the total number of protons and neutrons, \(i.\,e.\,\,A = Z + N\)
PHXII13:NUCLEI
363841
If the radius of a nucleus of mass number 3 is \(R\), then the radius of a nucleus of mass number 81 is
363842
The radius of the nucleus is proportional to, (if \(A\) is the atomic mass number)
1 \(A\)
2 \({A^3}\)
3 \({A^{1/3}}\)
4 \({A^{2/3}}\)
Explanation:
\(R = {R_0}{A^{1/3}}\)
PHXII13:NUCLEI
363843
Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio of radii of nucleus to that of Helium nucleus is \({(14)^{1/3}}\). The atomic number of nucleus will be
1 25
2 26
3 56
4 30
Explanation:
The radius of a nucleus is \(R=R_{0} A^{1 / 3}\) where \(R_{0}\) is a constant and \(A\) be its mass number. \(\therefore \quad \frac{R}{{{R_{He}}}} = {\left( {\frac{A}{4}} \right)^{1/3}}\) As \(\frac{R}{{{R_{He}}}} = {(14)^{1/3}}\) (given) \(\therefore \quad(14)^{1 / 3}=\left(\dfrac{A}{4}\right)^{1 / 3} ; 14=\dfrac{A}{4}\) or \(A=56\) The atomic number of nucleus is \(Z=A-N=56-30=26\)
