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JEE Exam  >  JEE Tests  >  Chemistry for JEE Main & Advanced  >  Atomic Structure- 2 - JEE MCQ

Atomic Structure- 2 - JEE MCQ


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30 Questions MCQ Test Chemistry for JEE Main & Advanced - Atomic Structure- 2

Atomic Structure- 2 for JEE 2025 is part of Chemistry for JEE Main & Advanced preparation. The Atomic Structure- 2 questions and answers have been prepared according to the JEE exam syllabus.The Atomic Structure- 2 MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Atomic Structure- 2 below.
Solutions of Atomic Structure- 2 questions in English are available as part of our Chemistry for JEE Main & Advanced for JEE & Atomic Structure- 2 solutions in Hindi for Chemistry for JEE Main & Advanced course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Atomic Structure- 2 | 30 questions in 60 minutes | Mock test for JEE preparation | Free important questions MCQ to study Chemistry for JEE Main & Advanced for JEE Exam | Download free PDF with solutions
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Which of the following curve correctly represent the photocurrent as a function of distance between the point source and the small metal plate?

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Atomic Structure- 2 - Question 4
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Give the correct order of initials T (true) or F (false) for following statements.

(i) The ratio of the radii of the first three Bohr orbit of hydrogen atom is 1 : 8 : 27

(ii) The ratio of magnitude of Total Energy : Kinetic Energy : Potential Energy for any orbit is 1 : 1 : 2

(iii) The frequency of a green light is 6 × 1014 Hz, then its wavelength is 500 nm

(iv) The ratio of de-broglie wavelength of a 'H' atom, 'He' atom and CH4 molecule moving with equal kinetic energy is 4 : 2 : 1
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A photon of wavelength 12.4 nm is used to emit an electron from the ground state of He+. Calculate debroglie wavelength (in Angstrom) of emitted electron.

(Given: hc = 1240 nm.eV)

[Fill your answer by multiplying it with (273.6)1/2]
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To determine how many electrons can fit in the orbital for which n=3 and l=1, follow these steps:

  1. Identify the orbital type:

    • The azimuthal quantum number l=1 corresponds to a p-orbital.

  2. Determine the number of orbitals:

    • For l=1, there are 2l+1=3 orbitals (px,py,pz).

  3. Calculate the number of electrons:

    • Each orbital can hold a maximum of 2 electrons (with opposite spins).

    • Therefore, the total number of electrons is 3 orbitals × 2 electrons/orbital=6 electrons.

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To determine the total number of electrons that can be accommodated in all the orbitals with principal quantum number n=2 and azimuthal quantum number l=1, follow these steps:

  1. Identify the orbital type:

    • The azimuthal quantum number l=1 corresponds to p-orbitals.

  2. Determine the number of orbitals:

    • For l=1, the magnetic quantum number ml​ can take values −1,0,+1. This means there are 3 orbitals (px,py,pz​).

  3. Calculate the number of electrons:

    • Each orbital can hold a maximum of 2 electrons (with opposite spins).

    • Therefore, the total number of electrons is 3 orbitals × 2 electrons/orbital=6 electrons.

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