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Nuclei Test - 27

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Nuclei Test - 27
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  • Question 1
    1 / -0
    The atomic mass of $$_{8}O^{16}$$ is $$15.9949 amu$$. Mass of one neutron and one proton is $$2.016490 amu$$ and the mass of an electron is $$0.00055 amu$$. The binding energy per nucleon of oxygen atom is
    Solution
    The number of proton $$p=8$$ and number of neutron, $$n=16-8=8 $$
    Total number of nucleon $$N=16$$
    Combined mass of nuclei $$= $$ total mass of proton + total mass of neutron $$=8\times 2.016490=16.13192 $$ amu
    Mass defect $$\Delta m =16.13192-15.9949=0.13702 $$ amu
    Binding energy per nucleon $$=\Delta m c^2/N=(0.137\times 931.49)/16= 127.6141/16=7.975MeV$$   where $$ 1  amu =931.49/c^2  MeV$$


  • Question 2
    1 / -0
    The mass of proton is 1.0073 u and that of neutron is 1.0087 u (u $$=$$ atomic mass unit). The binding energy of  $$ ^{4}_{2}He$$ is (Given : helium nucleus mass $$\approx $$ 4.0015 u)
    Solution
    $$BE=\Delta m\times 931$$
    $$=[2(1.0087+1.0073)-4.0015]\times 931$$
    $$=28.4 MeV$$
  • Question 3
    1 / -0
    If the masses of deuterium and that of helium are 2.0140 amu and 4.0026 amu, respectively and that 22.4 MeV energy is liberated in the reaction $$_6^3Li + _1^2H \to _2^4H + _2^4He$$, has the mass of $$_3^6Li$$ is
    Solution
    $$Li_3+2_1H\rightarrow 4_2H+4_2He$$
    $$\Delta m=-(2\times 4\cdot 0026-2\cdot 0140-m)$$
    $$m-5\cdot 9912$$
    $$22\cdot 4=m-5\cdot 9912\times 931\cdot 5$$
    $$0\cdot 024=m-5\cdot 9912$$
    $$\Rightarrow m=6\cdot 0152 amu$$
  • Question 4
    1 / -0
    The mass of a $$^7_3Li $$ nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of $$^7_3Li $$  nucleus is nearly
    Solution
    $$\begin{aligned}\Delta m &=0.042 \\&=0.042+2\times 1.66 \times 10^{-27}\mathrm{~kg} \\\text { Binding energy }&=\Delta \mathrm{m} c^{2} \\\text { Binding energy per nucleon } \\=& \frac{\Delta m c^{2}}{7} \\&[\operatorname{mass number}  \text { of } l i=7]\end{aligned}$$
    $$\begin{array}{l}=\frac{0.042 \times 1.66\times 10^{-27} \times\left(3 \times 10^{8}\right)^{2}}{7} \mathrm{~J} \\\text { To convert the energy in ev from } \\\text { joule, we will divide it by } 1.6 \times 10^{-19} \\=\frac{0.042 \times 1.66 \times 10^{-27} \times\left(3 \times 10^{8}\right)^{2}}{7 \times 1.6 \times 10^{-19}}\mathrm{ev} \\=5.4 \times 10^{6}\mathrm{ev} \\\therefore \text { Energy }\operatorname{lost} =5.4\mathrm{Mev}\end{array}$$
  • Question 5
    1 / -0
    The binding energy per nucleon in deutorium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a
    helium nucleus the energy released in the fusion is -

    Solution

  • Question 6
    1 / -0
    To obtain an isotope of a given radioactive atom, the atom must emit
    Solution
    Emission of-particle decreases the atomic number of the product by two and each $$\beta$$ - emission increases the atomic number by one. So the atomic number remains the same and the mass number is different. after emitting one alpha and 2 beta.
  • Question 7
    1 / -0
    Atoms of same element having same atomic number but different mass number is
    Solution
    $$\begin{array}{l}\text { Atoms having same atomic number but different mass } \\\text { number is called Isotope. } \\\therefore \text { Isotope is correct }\end{array}$$
  • Question 8
    1 / -0
    Which of the following is true for the following isotopes of uranium

    U$$^{235}$$ and U$$^{238}$$?
    Solution
    Natural Uranium contains $$3$$ radioactive isotopes $${ U }^{ 234 },{ U }^{ 235 }$$ and $${ U }^{ 238 }$$
    $${ U }^{ 238 }$$ has a mass number
    (proton $$92+$$neutron $$146$$) $$=238$$
    $${ U }^{ 235 }$$ has mass number (proton $$92+$$ neutron$$143$$) $$=235$$
    Hence we can say that both contain the same number of protons and electrons but $${ U }^{ 238 }$$ contains three more neutrons than $${ U }^{ 235 }$$
  • Question 9
    1 / -0
    In the process of nuclear fusion
    Solution
    Nuclear fusion is a reaction in which two or more atomic nuclei come close enough to form one or more different atomic nuclei and subatomic particles (neutrons or protons)
    Inside the sun fusion reaction take place at very high temperature and enormous gravitational pressure.
  • Question 10
    1 / -0
    For a nuclear fusion process, suitable nuclei are
    Solution
    The process in which two or more light, nuclei are combined into a single nucleus with the release of tremendous amount of energy is called as nuclear fusion
    Hence light nuclei are suitable for the nuclear fusion process
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