147662
A free neutron decays to proton but a free proton does not decay to a neutron This is because.
1 neutron is a composite particle made of proton and electron whereas proton is a fundamental particle
2 neutron is an uncharged particle whereas proton is a charged particle
3 neutron has larger rest mass
4 None of the above
Explanation:
C A free neutron decays to proton but a free proton does not decay to neutron this is because neutron has a larger rest mass.
SCRA- 2013
NUCLEAR PHYSICS
147669
In radioactive decay process, the negatively charged emitted $\beta$-particle are
1 the electrons present inside the nucleus
2 the electrons produced as a result of the decay of neutrons inside the nucleus
3 the electrons produced as a result of collisions between atoms
4 the electrons orbiting around the nucleus
Explanation:
B Beta decay can involve the emission of either electrons or positrons. The electrons or positrons emitted in a $\beta$-decay do not exist inside the nucleus. They are only created at the time of emission, just as photons are created when an atom makes a transition from higher to a lower energy state. In negative $\beta$-decay a neutron in the nucleus is transformed into proton, an electron and an antineutrino. Hence in radioactive decay process, the negatively charged emitted $\beta$-particles are the electrons produced as a result of the decay of neutrons present inside the nucleus.
UPSEE - 2007
NUCLEAR PHYSICS
147673
The nucleus ${ }_{92} U^{234}$ splits exactly in half in a fission reaction in which two neutrons are released. The resultant nuclei are :
1 ${ }_{46} \mathrm{Pd}^{116}$
2 ${ }_{45} \mathrm{Rh}^{117}$
3 ${ }_{45} \mathrm{Rh}^{116}$
4 ${ }_{46} \mathrm{Pd}^{117}$
Explanation:
A The splitting of ${ }_{92} \mathrm{U}^{234}$ is as follows ${ }_{92} \mathrm{U}^{234} \rightarrow{ }_{46} \mathrm{X}^{116}+{ }_{46} \mathrm{X}^{116}+2{ }_{0} \mathrm{n}^{1}+$ energy So, the resultant nuclei is $46 \mathrm{X}^{116}$ as ${ }_{46} \mathrm{Pd}^{116}$
UPSEE - 2004
NUCLEAR PHYSICS
147675
In a series of radioactive decays, if a nucleus of mass number 180 and atomic number 72 decays into another nucleus of mass number 172 and atomic number 69, then the number of $\alpha$ and $\beta$ particles released respectively are
1 2,3
2 2,2
3 2,1
4 2,0
5 1,3
Explanation:
C Mass number of nucleus is decreased by 8 and atomic number is decreased by 3 . So, the number of $\alpha$-particles are two and number of $\beta$-particle is one. ${ }_{72}^{180} \mathrm{X} \rightarrow{ }_{69}^{172} \mathrm{Y}+2\left({ }_{2}^{4} \mathrm{He}\right)+{ }_{-1}^{0} \mathrm{e}$
Kerala CEE - 2015
NUCLEAR PHYSICS
147679
A radioactive decay can form an isotope to the original nucleus with the emission of particles
1 one $\alpha$ and four $\beta$
2 one $\alpha$ and two $\beta$
3 one $\alpha$ and one $\beta$
4 four $\alpha$ and one $\beta$
5 two $\alpha$ and one $\beta$
Explanation:
B As the emission of an alpha particle decreases the atomic number and mass number by 2 and 4 respectively while the emission of a beta particle increases the atomic number by 1 but mass number remains the same. So the emission of one alpha particle and two beta particles from the nucleus can form an isotope of the original nucleus.
147662
A free neutron decays to proton but a free proton does not decay to a neutron This is because.
1 neutron is a composite particle made of proton and electron whereas proton is a fundamental particle
2 neutron is an uncharged particle whereas proton is a charged particle
3 neutron has larger rest mass
4 None of the above
Explanation:
C A free neutron decays to proton but a free proton does not decay to neutron this is because neutron has a larger rest mass.
SCRA- 2013
NUCLEAR PHYSICS
147669
In radioactive decay process, the negatively charged emitted $\beta$-particle are
1 the electrons present inside the nucleus
2 the electrons produced as a result of the decay of neutrons inside the nucleus
3 the electrons produced as a result of collisions between atoms
4 the electrons orbiting around the nucleus
Explanation:
B Beta decay can involve the emission of either electrons or positrons. The electrons or positrons emitted in a $\beta$-decay do not exist inside the nucleus. They are only created at the time of emission, just as photons are created when an atom makes a transition from higher to a lower energy state. In negative $\beta$-decay a neutron in the nucleus is transformed into proton, an electron and an antineutrino. Hence in radioactive decay process, the negatively charged emitted $\beta$-particles are the electrons produced as a result of the decay of neutrons present inside the nucleus.
UPSEE - 2007
NUCLEAR PHYSICS
147673
The nucleus ${ }_{92} U^{234}$ splits exactly in half in a fission reaction in which two neutrons are released. The resultant nuclei are :
1 ${ }_{46} \mathrm{Pd}^{116}$
2 ${ }_{45} \mathrm{Rh}^{117}$
3 ${ }_{45} \mathrm{Rh}^{116}$
4 ${ }_{46} \mathrm{Pd}^{117}$
Explanation:
A The splitting of ${ }_{92} \mathrm{U}^{234}$ is as follows ${ }_{92} \mathrm{U}^{234} \rightarrow{ }_{46} \mathrm{X}^{116}+{ }_{46} \mathrm{X}^{116}+2{ }_{0} \mathrm{n}^{1}+$ energy So, the resultant nuclei is $46 \mathrm{X}^{116}$ as ${ }_{46} \mathrm{Pd}^{116}$
UPSEE - 2004
NUCLEAR PHYSICS
147675
In a series of radioactive decays, if a nucleus of mass number 180 and atomic number 72 decays into another nucleus of mass number 172 and atomic number 69, then the number of $\alpha$ and $\beta$ particles released respectively are
1 2,3
2 2,2
3 2,1
4 2,0
5 1,3
Explanation:
C Mass number of nucleus is decreased by 8 and atomic number is decreased by 3 . So, the number of $\alpha$-particles are two and number of $\beta$-particle is one. ${ }_{72}^{180} \mathrm{X} \rightarrow{ }_{69}^{172} \mathrm{Y}+2\left({ }_{2}^{4} \mathrm{He}\right)+{ }_{-1}^{0} \mathrm{e}$
Kerala CEE - 2015
NUCLEAR PHYSICS
147679
A radioactive decay can form an isotope to the original nucleus with the emission of particles
1 one $\alpha$ and four $\beta$
2 one $\alpha$ and two $\beta$
3 one $\alpha$ and one $\beta$
4 four $\alpha$ and one $\beta$
5 two $\alpha$ and one $\beta$
Explanation:
B As the emission of an alpha particle decreases the atomic number and mass number by 2 and 4 respectively while the emission of a beta particle increases the atomic number by 1 but mass number remains the same. So the emission of one alpha particle and two beta particles from the nucleus can form an isotope of the original nucleus.
147662
A free neutron decays to proton but a free proton does not decay to a neutron This is because.
1 neutron is a composite particle made of proton and electron whereas proton is a fundamental particle
2 neutron is an uncharged particle whereas proton is a charged particle
3 neutron has larger rest mass
4 None of the above
Explanation:
C A free neutron decays to proton but a free proton does not decay to neutron this is because neutron has a larger rest mass.
SCRA- 2013
NUCLEAR PHYSICS
147669
In radioactive decay process, the negatively charged emitted $\beta$-particle are
1 the electrons present inside the nucleus
2 the electrons produced as a result of the decay of neutrons inside the nucleus
3 the electrons produced as a result of collisions between atoms
4 the electrons orbiting around the nucleus
Explanation:
B Beta decay can involve the emission of either electrons or positrons. The electrons or positrons emitted in a $\beta$-decay do not exist inside the nucleus. They are only created at the time of emission, just as photons are created when an atom makes a transition from higher to a lower energy state. In negative $\beta$-decay a neutron in the nucleus is transformed into proton, an electron and an antineutrino. Hence in radioactive decay process, the negatively charged emitted $\beta$-particles are the electrons produced as a result of the decay of neutrons present inside the nucleus.
UPSEE - 2007
NUCLEAR PHYSICS
147673
The nucleus ${ }_{92} U^{234}$ splits exactly in half in a fission reaction in which two neutrons are released. The resultant nuclei are :
1 ${ }_{46} \mathrm{Pd}^{116}$
2 ${ }_{45} \mathrm{Rh}^{117}$
3 ${ }_{45} \mathrm{Rh}^{116}$
4 ${ }_{46} \mathrm{Pd}^{117}$
Explanation:
A The splitting of ${ }_{92} \mathrm{U}^{234}$ is as follows ${ }_{92} \mathrm{U}^{234} \rightarrow{ }_{46} \mathrm{X}^{116}+{ }_{46} \mathrm{X}^{116}+2{ }_{0} \mathrm{n}^{1}+$ energy So, the resultant nuclei is $46 \mathrm{X}^{116}$ as ${ }_{46} \mathrm{Pd}^{116}$
UPSEE - 2004
NUCLEAR PHYSICS
147675
In a series of radioactive decays, if a nucleus of mass number 180 and atomic number 72 decays into another nucleus of mass number 172 and atomic number 69, then the number of $\alpha$ and $\beta$ particles released respectively are
1 2,3
2 2,2
3 2,1
4 2,0
5 1,3
Explanation:
C Mass number of nucleus is decreased by 8 and atomic number is decreased by 3 . So, the number of $\alpha$-particles are two and number of $\beta$-particle is one. ${ }_{72}^{180} \mathrm{X} \rightarrow{ }_{69}^{172} \mathrm{Y}+2\left({ }_{2}^{4} \mathrm{He}\right)+{ }_{-1}^{0} \mathrm{e}$
Kerala CEE - 2015
NUCLEAR PHYSICS
147679
A radioactive decay can form an isotope to the original nucleus with the emission of particles
1 one $\alpha$ and four $\beta$
2 one $\alpha$ and two $\beta$
3 one $\alpha$ and one $\beta$
4 four $\alpha$ and one $\beta$
5 two $\alpha$ and one $\beta$
Explanation:
B As the emission of an alpha particle decreases the atomic number and mass number by 2 and 4 respectively while the emission of a beta particle increases the atomic number by 1 but mass number remains the same. So the emission of one alpha particle and two beta particles from the nucleus can form an isotope of the original nucleus.
147662
A free neutron decays to proton but a free proton does not decay to a neutron This is because.
1 neutron is a composite particle made of proton and electron whereas proton is a fundamental particle
2 neutron is an uncharged particle whereas proton is a charged particle
3 neutron has larger rest mass
4 None of the above
Explanation:
C A free neutron decays to proton but a free proton does not decay to neutron this is because neutron has a larger rest mass.
SCRA- 2013
NUCLEAR PHYSICS
147669
In radioactive decay process, the negatively charged emitted $\beta$-particle are
1 the electrons present inside the nucleus
2 the electrons produced as a result of the decay of neutrons inside the nucleus
3 the electrons produced as a result of collisions between atoms
4 the electrons orbiting around the nucleus
Explanation:
B Beta decay can involve the emission of either electrons or positrons. The electrons or positrons emitted in a $\beta$-decay do not exist inside the nucleus. They are only created at the time of emission, just as photons are created when an atom makes a transition from higher to a lower energy state. In negative $\beta$-decay a neutron in the nucleus is transformed into proton, an electron and an antineutrino. Hence in radioactive decay process, the negatively charged emitted $\beta$-particles are the electrons produced as a result of the decay of neutrons present inside the nucleus.
UPSEE - 2007
NUCLEAR PHYSICS
147673
The nucleus ${ }_{92} U^{234}$ splits exactly in half in a fission reaction in which two neutrons are released. The resultant nuclei are :
1 ${ }_{46} \mathrm{Pd}^{116}$
2 ${ }_{45} \mathrm{Rh}^{117}$
3 ${ }_{45} \mathrm{Rh}^{116}$
4 ${ }_{46} \mathrm{Pd}^{117}$
Explanation:
A The splitting of ${ }_{92} \mathrm{U}^{234}$ is as follows ${ }_{92} \mathrm{U}^{234} \rightarrow{ }_{46} \mathrm{X}^{116}+{ }_{46} \mathrm{X}^{116}+2{ }_{0} \mathrm{n}^{1}+$ energy So, the resultant nuclei is $46 \mathrm{X}^{116}$ as ${ }_{46} \mathrm{Pd}^{116}$
UPSEE - 2004
NUCLEAR PHYSICS
147675
In a series of radioactive decays, if a nucleus of mass number 180 and atomic number 72 decays into another nucleus of mass number 172 and atomic number 69, then the number of $\alpha$ and $\beta$ particles released respectively are
1 2,3
2 2,2
3 2,1
4 2,0
5 1,3
Explanation:
C Mass number of nucleus is decreased by 8 and atomic number is decreased by 3 . So, the number of $\alpha$-particles are two and number of $\beta$-particle is one. ${ }_{72}^{180} \mathrm{X} \rightarrow{ }_{69}^{172} \mathrm{Y}+2\left({ }_{2}^{4} \mathrm{He}\right)+{ }_{-1}^{0} \mathrm{e}$
Kerala CEE - 2015
NUCLEAR PHYSICS
147679
A radioactive decay can form an isotope to the original nucleus with the emission of particles
1 one $\alpha$ and four $\beta$
2 one $\alpha$ and two $\beta$
3 one $\alpha$ and one $\beta$
4 four $\alpha$ and one $\beta$
5 two $\alpha$ and one $\beta$
Explanation:
B As the emission of an alpha particle decreases the atomic number and mass number by 2 and 4 respectively while the emission of a beta particle increases the atomic number by 1 but mass number remains the same. So the emission of one alpha particle and two beta particles from the nucleus can form an isotope of the original nucleus.
147662
A free neutron decays to proton but a free proton does not decay to a neutron This is because.
1 neutron is a composite particle made of proton and electron whereas proton is a fundamental particle
2 neutron is an uncharged particle whereas proton is a charged particle
3 neutron has larger rest mass
4 None of the above
Explanation:
C A free neutron decays to proton but a free proton does not decay to neutron this is because neutron has a larger rest mass.
SCRA- 2013
NUCLEAR PHYSICS
147669
In radioactive decay process, the negatively charged emitted $\beta$-particle are
1 the electrons present inside the nucleus
2 the electrons produced as a result of the decay of neutrons inside the nucleus
3 the electrons produced as a result of collisions between atoms
4 the electrons orbiting around the nucleus
Explanation:
B Beta decay can involve the emission of either electrons or positrons. The electrons or positrons emitted in a $\beta$-decay do not exist inside the nucleus. They are only created at the time of emission, just as photons are created when an atom makes a transition from higher to a lower energy state. In negative $\beta$-decay a neutron in the nucleus is transformed into proton, an electron and an antineutrino. Hence in radioactive decay process, the negatively charged emitted $\beta$-particles are the electrons produced as a result of the decay of neutrons present inside the nucleus.
UPSEE - 2007
NUCLEAR PHYSICS
147673
The nucleus ${ }_{92} U^{234}$ splits exactly in half in a fission reaction in which two neutrons are released. The resultant nuclei are :
1 ${ }_{46} \mathrm{Pd}^{116}$
2 ${ }_{45} \mathrm{Rh}^{117}$
3 ${ }_{45} \mathrm{Rh}^{116}$
4 ${ }_{46} \mathrm{Pd}^{117}$
Explanation:
A The splitting of ${ }_{92} \mathrm{U}^{234}$ is as follows ${ }_{92} \mathrm{U}^{234} \rightarrow{ }_{46} \mathrm{X}^{116}+{ }_{46} \mathrm{X}^{116}+2{ }_{0} \mathrm{n}^{1}+$ energy So, the resultant nuclei is $46 \mathrm{X}^{116}$ as ${ }_{46} \mathrm{Pd}^{116}$
UPSEE - 2004
NUCLEAR PHYSICS
147675
In a series of radioactive decays, if a nucleus of mass number 180 and atomic number 72 decays into another nucleus of mass number 172 and atomic number 69, then the number of $\alpha$ and $\beta$ particles released respectively are
1 2,3
2 2,2
3 2,1
4 2,0
5 1,3
Explanation:
C Mass number of nucleus is decreased by 8 and atomic number is decreased by 3 . So, the number of $\alpha$-particles are two and number of $\beta$-particle is one. ${ }_{72}^{180} \mathrm{X} \rightarrow{ }_{69}^{172} \mathrm{Y}+2\left({ }_{2}^{4} \mathrm{He}\right)+{ }_{-1}^{0} \mathrm{e}$
Kerala CEE - 2015
NUCLEAR PHYSICS
147679
A radioactive decay can form an isotope to the original nucleus with the emission of particles
1 one $\alpha$ and four $\beta$
2 one $\alpha$ and two $\beta$
3 one $\alpha$ and one $\beta$
4 four $\alpha$ and one $\beta$
5 two $\alpha$ and one $\beta$
Explanation:
B As the emission of an alpha particle decreases the atomic number and mass number by 2 and 4 respectively while the emission of a beta particle increases the atomic number by 1 but mass number remains the same. So the emission of one alpha particle and two beta particles from the nucleus can form an isotope of the original nucleus.
