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Structure of the Atoms Test - 24

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Structure of the Atoms Test - 24
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  • Question 1
    1 / -0
    The total number of neutrons in all isotopes of hydrogen is equal to:
    Solution
    Hydrogen has 3 isotopes.

                   $$_{1}^{1}\textrm{H}  $$      $$_{1}^{2}\textrm{H}$$          $$_{1}^{3}\textrm{H}$$
    neutron  0 
           1
             2
    proton    1
           1   
             1
     
    The total number of neutrons in all the isotopes of Hydrogen $$=0+1+2=3$$.
  • Question 2
    1 / -0
    In Rutherford's experiment, most of the alpha particles go straight through the foil because
    Solution
    The observation that most alpha particles passed straight through the gold foil led Rutherford to conclude that the positive charge in an atom is concentrated in a very small area that is called nucleus. Atoms have mostly empty space. Electrons, which are negatively charged, are distributed throughout this space but take up a very small part of it.
  • Question 3
    1 / -0
    Which of the following point is not shown by the Rutherford alpha scattering experiment?
    Solution
    The conclusions of Rutherford's alpha scattering experiment do not include that the positively charged particles move with great velocities. The positively charged particles are present at the center of the atom called a nucleus.
  • Question 4
    1 / -0
    In Rutherford experiment, most of the alpha particles go straight through the foil because:
    Solution
    Rutherford's model describes atom as a tiny, dense, positively charged core called the nucleus in which almost the entire mass is concentrated, around which electrons circle around at a certain distance. In his experiment, he bombarded a thin gold foil with alpha particles. 
    Rutherford reasoned that since most of the atom is empty, the alpha particles passed straight through the gold foil. 
    So, option C is correct.
  • Question 5
    1 / -0
    In Rutherford's $$\alpha$$-rays scattering experiment, gold foils are used because of ___________.
    Solution
    Malleability is the material's ability to form thin sheets. In Rutherford's ray scattering experiment, gold foil is used because of its high malleability. Very thin gold foil is used in the experiment and gold is capable of being rolled into extremely thin foils.
  • Question 6
    1 / -0
    Which of the following properties are similar for isotopes?
    Solution
    Isotopes have same atomic numbers but different mass numbers. This also means that the isotopes have the same number of electrons. 
    Chemical properties are determined by electronic configuration. So, with the same electronic configuration, they have similar chemical properties. 
  • Question 7
    1 / -0
    To which of the following is Bohr's theory applicable:
    I) $$He^{+}$$ 
    II) $$Li^{2+}$$
    III) Tritium  
    IV) $$Be^{2+}$$
    The correct combination is:
    Solution
    Bohr's atomic model explains the spectra of single-electron atoms like $$H$$-atom. It does not explain the spectra of multi-electron atoms. 

    So the atoms which are isoelectronic and have the same electronic configuration are well explained by this atomic model.

    $$He^+$$, $$Li^{2+}$$ and tritium $$_{ 1 }{ H }^{ 3 }$$ have same number of electrons as that of hydrogen atom i.e. $$1$$ while $$Be^{2+}$$ is formed by losing $$2e$$. So, it is a two-electron system.

    Option D is correct.
  • Question 8
    1 / -0
    Which of the following statement is true about Rutherford’s experiment?
    Solution
    According to Rutherford's experiment, $$\alpha-$$ particles or helium nuclei impinged on a metal foil and got scattered. 

    Hence, the correct option is D.
  • Question 9
    1 / -0
    In Rutherford's alpha-ray scattering experiment, a screen is used to detect the alpha particles which is coated by:
    Solution
    In Rutherford's alpha-ray scattering experiment, the alpha particles are detected using a screen coated with zinc sulphide.
  • Question 10
    1 / -0
    The electron distribution in an aluminium atom is:
    Solution
    Atomic number of Al = 13. 
    K, L, M shell will accommodate 2, 8, 3 electrons ($$2n^2$$ rule).  Therefore, the electronic configuration is 2, 8, 3.
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