Atoms

Atoms

Easy Questions

Modern Physics

Dual Nature of Radiation and Matter
Atoms
Nuclei
Semiconductor Electronics : Materials, Devices and Simple Circuits
  1. The total energy of an electron in the first excited state of hydrogen atom is about –3.4 eV. Its kinetic energy in this state is ​​








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    K.E =
    Z2
    		(13.6 eV)
    n2

    Mechanical energy =
    - Z2
    		(13.6 eV)
    n2

    ∴ K.E. in 2nd orbital for hydrogen ​= – Mechanical energy
    =
    12
    		(13.6) = + 3.4 eV
    22

    Correct Option: A

    K.E =
    Z2
    		(13.6 eV)
    n2

    Mechanical energy =
    - Z2
    		(13.6 eV)
    n2

    ∴ K.E. in 2nd orbital for hydrogen ​= – Mechanical energy
    =
    12
    		(13.6) = + 3.4 eV
    22

  1. Ionization potential of hydrogen atom is 13.6eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. According to Bohr’s theory, the spectral lines emitted by hydrogen will be








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    Energy of ground state 13.6 eV ​Energy of first excited state

    = -
    13.6
    		= - 3.4eV
    4

    Energy of second excited state
    = -
    13.6
    		= - 1.5eV
    9

    Difference between ground state and 2nd excited state = 13.6 – 1.5 = 12.1 eV ​
    So, electron can be excited upto 3rd orbit ​No. of possible transition ​
    1 → 2, 1 → 3, 2 → 3 ​
    So, three lines are possible.

    Correct Option: A

    Energy of ground state 13.6 eV ​Energy of first excited state

    = -
    13.6
    		= - 3.4eV
    4

    Energy of second excited state
    = -
    13.6
    		= - 1.5eV
    9

    Difference between ground state and 2nd excited state = 13.6 – 1.5 = 12.1 eV ​
    So, electron can be excited upto 3rd orbit ​No. of possible transition ​
    1 → 2, 1 → 3, 2 → 3 ​
    So, three lines are possible.

  1. The total energy of electron in the ground state of hydrogen atom is – 13.6 eV. The kinetic energy of an electron in the first excited state is ​​








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    Energy in the first excited state

    =
    - 13.6
    n2

    =
    - 13.6
    		= - 3.4eV
    22

    But K.E. = –(Total energy) = +3.4 eV.

    Correct Option: D

    Energy in the first excited state

    =
    - 13.6
    n2

    =
    - 13.6
    		= - 3.4eV
    22

    But K.E. = –(Total energy) = +3.4 eV.

  1. The ground state energy of hydrogen atom is 13.6eV. When its electron is in the first excited state, its excitation energy is








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    When the electron is in first excited state (n = 2), the excitation energy is given by

    ∆E = E2 – E1
    We have,

    En = -
    13.6
    		eV
    n2

    ∴ E2 = -
    13.6
    		= - 3.4eV
    22

    Given E1 = –13.6eV ​
    ∴ ∆ E = ( – 3.4) – ( – 13.6) = 10.2 eV.

    Correct Option: C

    When the electron is in first excited state (n = 2), the excitation energy is given by

    ∆E = E2 – E1
    We have,

    En = -
    13.6
    		eV
    n2

    ∴ E2 = -
    13.6
    		= - 3.4eV
    22

    Given E1 = –13.6eV ​
    ∴ ∆ E = ( – 3.4) – ( – 13.6) = 10.2 eV.