147457
What is the binding energy of ${ }_{14}^{29} \mathrm{Si}$, whose atomic mass is $28.976495 \mathrm{u}$ ? Mass of proton $=1.007276 \mathrm{u}$ Mass of neutron $=\mathbf{1 . 0 0 8 6 6 4} \mathrm{u}$ (neglect the electron mass) (assume, $1 \mathrm{u}=931.5 \mathrm{MeV}$ )
1 $237.84 \mathrm{MeV}$
2 $421.72 \mathrm{MeV}$
3 $387.21 \mathrm{MeV}$
4 $116.35 \mathrm{MeV}$
Explanation:
A Given that, Mass of the ${ }_{14} \mathrm{Si}^{29}=28.976495 \mathrm{u}$ Number of Proton $=14$ Then, mass of 14 proton $=14 \times$ mass of one proton $=14 \times 1.007276 \mathrm{u}$ $=14.101864 \mathrm{u}$ Number of neutrons $=$ Atomic mass - Number of Proton $=29-14=15$ Mass of 15 neutron $=15 \times$ mass of 1 neutron $=15 \times 1.008664 \mathrm{u}$ $=15.12996 \mathrm{u}$ Total mass of constituents of $\mathrm{Si}$ atom $=14.101864+15.12996$ $=29.231824 \mathrm{u}$ So, the mass defect $=29.231824-28.976495$ $=0.255329 \mathrm{u}$ B.E $=0.255329 \times 931.5$ $=237.84 \mathrm{MeV}$
TS- EAMCET-09.09.2020
NUCLEAR PHYSICS
147458
The binding energy (BE) per nucleon for an element is $7.14 \mathrm{MeV}$. If the BE of element is 28.6 MeV, then the number of nucleons in the element is
1 4
2 8
3 16
4 32
Explanation:
A Given that, Binding energy per nucleon $=7.14 \mathrm{MeV}$ Binding energy of element $=28.6 \mathrm{MeV}$ Number of Nucleons $=\frac{\text { Binding energy of element }}{\text { Binding energy of per nucleon }}$ $\mathrm{n}=\frac{28.6}{7.14}$ $\mathrm{n}=4$
TS- EAMCET.11.09.2020
NUCLEAR PHYSICS
147459
Binding energy per nucleon plot against the mass number for stable nuclei is shown in the figure. Which curve is correct?
1 $\mathrm{A}$
2 $\mathrm{B}$
3 $\mathrm{C}$
4 $\mathrm{D}$
Explanation:
C Binding energy per nucleon $=\frac{\text { Total binding energy }}{\text { Number of Nucleon }}$
Manipal UGET-2019
NUCLEAR PHYSICS
147461
The energy equivalent of $0.5 \mathrm{~g}$ of a substance is
1 $4.5 \times 10^{13} \mathrm{~J}$
2 $1.5 \times 10^{13} \mathrm{~J}$
3 $0.5 \times 10^{13} \mathrm{~J}$
4 $4.5 \times 10^{16} \mathrm{~J}$
Explanation:
A Given that, Mass $(\mathrm{m})=0.5 \mathrm{~g}=0.5 \times 10^{-3} \mathrm{~kg}$ Speed of light $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{sec}$ So energy $(E)=(\Delta \mathrm{m}) \mathrm{c}^{2}$ $\mathrm{E}=0.5 \times 10^{-3} \times\left(3 \times 10^{8}\right)^{2}$ $\mathrm{E}=4.5 \times 10^{13} \text { Joule. }$
147457
What is the binding energy of ${ }_{14}^{29} \mathrm{Si}$, whose atomic mass is $28.976495 \mathrm{u}$ ? Mass of proton $=1.007276 \mathrm{u}$ Mass of neutron $=\mathbf{1 . 0 0 8 6 6 4} \mathrm{u}$ (neglect the electron mass) (assume, $1 \mathrm{u}=931.5 \mathrm{MeV}$ )
1 $237.84 \mathrm{MeV}$
2 $421.72 \mathrm{MeV}$
3 $387.21 \mathrm{MeV}$
4 $116.35 \mathrm{MeV}$
Explanation:
A Given that, Mass of the ${ }_{14} \mathrm{Si}^{29}=28.976495 \mathrm{u}$ Number of Proton $=14$ Then, mass of 14 proton $=14 \times$ mass of one proton $=14 \times 1.007276 \mathrm{u}$ $=14.101864 \mathrm{u}$ Number of neutrons $=$ Atomic mass - Number of Proton $=29-14=15$ Mass of 15 neutron $=15 \times$ mass of 1 neutron $=15 \times 1.008664 \mathrm{u}$ $=15.12996 \mathrm{u}$ Total mass of constituents of $\mathrm{Si}$ atom $=14.101864+15.12996$ $=29.231824 \mathrm{u}$ So, the mass defect $=29.231824-28.976495$ $=0.255329 \mathrm{u}$ B.E $=0.255329 \times 931.5$ $=237.84 \mathrm{MeV}$
TS- EAMCET-09.09.2020
NUCLEAR PHYSICS
147458
The binding energy (BE) per nucleon for an element is $7.14 \mathrm{MeV}$. If the BE of element is 28.6 MeV, then the number of nucleons in the element is
1 4
2 8
3 16
4 32
Explanation:
A Given that, Binding energy per nucleon $=7.14 \mathrm{MeV}$ Binding energy of element $=28.6 \mathrm{MeV}$ Number of Nucleons $=\frac{\text { Binding energy of element }}{\text { Binding energy of per nucleon }}$ $\mathrm{n}=\frac{28.6}{7.14}$ $\mathrm{n}=4$
TS- EAMCET.11.09.2020
NUCLEAR PHYSICS
147459
Binding energy per nucleon plot against the mass number for stable nuclei is shown in the figure. Which curve is correct?
1 $\mathrm{A}$
2 $\mathrm{B}$
3 $\mathrm{C}$
4 $\mathrm{D}$
Explanation:
C Binding energy per nucleon $=\frac{\text { Total binding energy }}{\text { Number of Nucleon }}$
Manipal UGET-2019
NUCLEAR PHYSICS
147461
The energy equivalent of $0.5 \mathrm{~g}$ of a substance is
1 $4.5 \times 10^{13} \mathrm{~J}$
2 $1.5 \times 10^{13} \mathrm{~J}$
3 $0.5 \times 10^{13} \mathrm{~J}$
4 $4.5 \times 10^{16} \mathrm{~J}$
Explanation:
A Given that, Mass $(\mathrm{m})=0.5 \mathrm{~g}=0.5 \times 10^{-3} \mathrm{~kg}$ Speed of light $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{sec}$ So energy $(E)=(\Delta \mathrm{m}) \mathrm{c}^{2}$ $\mathrm{E}=0.5 \times 10^{-3} \times\left(3 \times 10^{8}\right)^{2}$ $\mathrm{E}=4.5 \times 10^{13} \text { Joule. }$
147457
What is the binding energy of ${ }_{14}^{29} \mathrm{Si}$, whose atomic mass is $28.976495 \mathrm{u}$ ? Mass of proton $=1.007276 \mathrm{u}$ Mass of neutron $=\mathbf{1 . 0 0 8 6 6 4} \mathrm{u}$ (neglect the electron mass) (assume, $1 \mathrm{u}=931.5 \mathrm{MeV}$ )
1 $237.84 \mathrm{MeV}$
2 $421.72 \mathrm{MeV}$
3 $387.21 \mathrm{MeV}$
4 $116.35 \mathrm{MeV}$
Explanation:
A Given that, Mass of the ${ }_{14} \mathrm{Si}^{29}=28.976495 \mathrm{u}$ Number of Proton $=14$ Then, mass of 14 proton $=14 \times$ mass of one proton $=14 \times 1.007276 \mathrm{u}$ $=14.101864 \mathrm{u}$ Number of neutrons $=$ Atomic mass - Number of Proton $=29-14=15$ Mass of 15 neutron $=15 \times$ mass of 1 neutron $=15 \times 1.008664 \mathrm{u}$ $=15.12996 \mathrm{u}$ Total mass of constituents of $\mathrm{Si}$ atom $=14.101864+15.12996$ $=29.231824 \mathrm{u}$ So, the mass defect $=29.231824-28.976495$ $=0.255329 \mathrm{u}$ B.E $=0.255329 \times 931.5$ $=237.84 \mathrm{MeV}$
TS- EAMCET-09.09.2020
NUCLEAR PHYSICS
147458
The binding energy (BE) per nucleon for an element is $7.14 \mathrm{MeV}$. If the BE of element is 28.6 MeV, then the number of nucleons in the element is
1 4
2 8
3 16
4 32
Explanation:
A Given that, Binding energy per nucleon $=7.14 \mathrm{MeV}$ Binding energy of element $=28.6 \mathrm{MeV}$ Number of Nucleons $=\frac{\text { Binding energy of element }}{\text { Binding energy of per nucleon }}$ $\mathrm{n}=\frac{28.6}{7.14}$ $\mathrm{n}=4$
TS- EAMCET.11.09.2020
NUCLEAR PHYSICS
147459
Binding energy per nucleon plot against the mass number for stable nuclei is shown in the figure. Which curve is correct?
1 $\mathrm{A}$
2 $\mathrm{B}$
3 $\mathrm{C}$
4 $\mathrm{D}$
Explanation:
C Binding energy per nucleon $=\frac{\text { Total binding energy }}{\text { Number of Nucleon }}$
Manipal UGET-2019
NUCLEAR PHYSICS
147461
The energy equivalent of $0.5 \mathrm{~g}$ of a substance is
1 $4.5 \times 10^{13} \mathrm{~J}$
2 $1.5 \times 10^{13} \mathrm{~J}$
3 $0.5 \times 10^{13} \mathrm{~J}$
4 $4.5 \times 10^{16} \mathrm{~J}$
Explanation:
A Given that, Mass $(\mathrm{m})=0.5 \mathrm{~g}=0.5 \times 10^{-3} \mathrm{~kg}$ Speed of light $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{sec}$ So energy $(E)=(\Delta \mathrm{m}) \mathrm{c}^{2}$ $\mathrm{E}=0.5 \times 10^{-3} \times\left(3 \times 10^{8}\right)^{2}$ $\mathrm{E}=4.5 \times 10^{13} \text { Joule. }$
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NUCLEAR PHYSICS
147457
What is the binding energy of ${ }_{14}^{29} \mathrm{Si}$, whose atomic mass is $28.976495 \mathrm{u}$ ? Mass of proton $=1.007276 \mathrm{u}$ Mass of neutron $=\mathbf{1 . 0 0 8 6 6 4} \mathrm{u}$ (neglect the electron mass) (assume, $1 \mathrm{u}=931.5 \mathrm{MeV}$ )
1 $237.84 \mathrm{MeV}$
2 $421.72 \mathrm{MeV}$
3 $387.21 \mathrm{MeV}$
4 $116.35 \mathrm{MeV}$
Explanation:
A Given that, Mass of the ${ }_{14} \mathrm{Si}^{29}=28.976495 \mathrm{u}$ Number of Proton $=14$ Then, mass of 14 proton $=14 \times$ mass of one proton $=14 \times 1.007276 \mathrm{u}$ $=14.101864 \mathrm{u}$ Number of neutrons $=$ Atomic mass - Number of Proton $=29-14=15$ Mass of 15 neutron $=15 \times$ mass of 1 neutron $=15 \times 1.008664 \mathrm{u}$ $=15.12996 \mathrm{u}$ Total mass of constituents of $\mathrm{Si}$ atom $=14.101864+15.12996$ $=29.231824 \mathrm{u}$ So, the mass defect $=29.231824-28.976495$ $=0.255329 \mathrm{u}$ B.E $=0.255329 \times 931.5$ $=237.84 \mathrm{MeV}$
TS- EAMCET-09.09.2020
NUCLEAR PHYSICS
147458
The binding energy (BE) per nucleon for an element is $7.14 \mathrm{MeV}$. If the BE of element is 28.6 MeV, then the number of nucleons in the element is
1 4
2 8
3 16
4 32
Explanation:
A Given that, Binding energy per nucleon $=7.14 \mathrm{MeV}$ Binding energy of element $=28.6 \mathrm{MeV}$ Number of Nucleons $=\frac{\text { Binding energy of element }}{\text { Binding energy of per nucleon }}$ $\mathrm{n}=\frac{28.6}{7.14}$ $\mathrm{n}=4$
TS- EAMCET.11.09.2020
NUCLEAR PHYSICS
147459
Binding energy per nucleon plot against the mass number for stable nuclei is shown in the figure. Which curve is correct?
1 $\mathrm{A}$
2 $\mathrm{B}$
3 $\mathrm{C}$
4 $\mathrm{D}$
Explanation:
C Binding energy per nucleon $=\frac{\text { Total binding energy }}{\text { Number of Nucleon }}$
Manipal UGET-2019
NUCLEAR PHYSICS
147461
The energy equivalent of $0.5 \mathrm{~g}$ of a substance is
1 $4.5 \times 10^{13} \mathrm{~J}$
2 $1.5 \times 10^{13} \mathrm{~J}$
3 $0.5 \times 10^{13} \mathrm{~J}$
4 $4.5 \times 10^{16} \mathrm{~J}$
Explanation:
A Given that, Mass $(\mathrm{m})=0.5 \mathrm{~g}=0.5 \times 10^{-3} \mathrm{~kg}$ Speed of light $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{sec}$ So energy $(E)=(\Delta \mathrm{m}) \mathrm{c}^{2}$ $\mathrm{E}=0.5 \times 10^{-3} \times\left(3 \times 10^{8}\right)^{2}$ $\mathrm{E}=4.5 \times 10^{13} \text { Joule. }$
