Self Studies

Structure of Atom Test - 9

Result Self Studies

Structure of Atom Test - 9
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0

    Find energy of the photons which correspond to light of frequency 3×1015 Hz (Hint: h = Planck’s constant = 6.626 × 10−34 Js)

    Solution

    We know Planck's equation is

    E = hν

    where E is energy, h is Planck’s constant, and ν is frequency.

    Put given values,

    E = 6.626 × 10−34 ×  3 × 1015=1.988×10−18J

  • Question 2
    1 / -0

    Lines in the hydrogen spectrum which appear in the ultraviolet region of the electromagnetic spectrum, then they are called as

    Solution

    The Lyman series is a hydrogen spectral series of transitions and resulting ultraviolet emission lines of the hydrogen atom as an electron goes from n ≥ 2 to n = 1 (where n is the principal quantum number), the lowest energy level of the electron.

  • Question 3
    1 / -0

    Lines in the hydrogen spectrum which appear in the infrared region of the electromagnetic Spectrum, then they are called as

    Solution

    The Lyman series lies in the ultraviolet, whereas the Paschen, Brackett, and Pfund series lie in the infrared.

  • Question 4
    1 / -0

    Lines in the hydrogen spectrum which appear in the visible region of the electromagnetic Spectrum, then they are called as

    Solution

    The Balmer series or Balmer lines in atomic physics, is the designation of one of a set of six named series describing the spectral line emissions of the hydrogen atom. The Balmer series is calculated using the Balmer formula, an empirical equation discovered by Johann Balmer in 1885 Balmar series lies in visible region of Electromagnetic spectrum.

  • Question 5
    1 / -0

    How many protons, neutrons and electrons are in 56Fe3+?

    Solution

    Atomic number of iron (Fe) = 26

    So number of protons = 26

    Number of protons = number of electrons

    But To gain 3+ charge it should loose 3 electron. So number of electrons present = 26-3 = 23

    Now atomic mass = 56 (Given).

    Number of neutrons = atomic mass - number of protons = 56-26 = 30

  • Question 6
    1 / -0

    Give the number of electrons in the species, O2 and O+2.

    Solution

    atomic number O has atomic number = 8 so number of electrons in O2 = 16

    while in O+2 there is one unit positive charge so no. of electron =15.

  • Question 7
    1 / -0

    An atom of an element contains 29 electrons and 35 neutrons. The number of protons are:

    Solution

    In a atom no. of protons = no. of electrons i.e. P = E while this is not true in case of ions.

    So number of protons in given atom is 29.

Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now