JEE Exam  >  JEE Tests  >  Daily Test for JEE Preparation  >  Test: Bohr’s Model of Atom (April 21) - JEE MCQ

Test: Bohr’s Model of Atom (April 21) - JEE MCQ


Test Description

10 Questions MCQ Test Daily Test for JEE Preparation - Test: Bohr’s Model of Atom (April 21)

Test: Bohr’s Model of Atom (April 21) for JEE 2024 is part of Daily Test for JEE Preparation preparation. The Test: Bohr’s Model of Atom (April 21) questions and answers have been prepared according to the JEE exam syllabus.The Test: Bohr’s Model of Atom (April 21) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Bohr’s Model of Atom (April 21) below.
Solutions of Test: Bohr’s Model of Atom (April 21) questions in English are available as part of our Daily Test for JEE Preparation for JEE & Test: Bohr’s Model of Atom (April 21) solutions in Hindi for Daily Test for JEE Preparation course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Test: Bohr’s Model of Atom (April 21) | 10 questions in 20 minutes | Mock test for JEE preparation | Free important questions MCQ to study Daily Test for JEE Preparation for JEE Exam | Download free PDF with solutions
Test: Bohr’s Model of Atom (April 21) - Question 1

Wave number of a spectral line for a given transition is x cm-1 for He+, then its value for Be3+ (isoelectronic of He+)for the same transition is

Detailed Solution for Test: Bohr’s Model of Atom (April 21) - Question 1



Test: Bohr’s Model of Atom (April 21) - Question 2

An electron in H-atom in its ground state absorbs 1.5 times as much as energy as the minimum required for its escape from the atom

Q.

Thus, kinetic energy given to the emitted electron is

Detailed Solution for Test: Bohr’s Model of Atom (April 21) - Question 2

E1 = Energy of H-atom in the ground state = 13.6 eV 

Energy absorbed = (13.6 x 1.5) = 20.4 eV
E2 = Energy of the excited state
= 13.6+ 20.4= 34.0 eV
ΔE = KE = (E2 - E1)
= 34.0 - 13.6 = 20.4 eV

1 Crore+ students have signed up on EduRev. Have you? Download the App
Test: Bohr’s Model of Atom (April 21) - Question 3

Ionisation energy of He+ is 19.6x10-18 J atom -1. The energy of the first stationary state (n = 1)of Li2+ is

Detailed Solution for Test: Bohr’s Model of Atom (April 21) - Question 3

Ionisation energy = - Energy of the electron






Test: Bohr’s Model of Atom (April 21) - Question 4

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is (a0 is Bohr radius)

[AIEEE 2012]

Detailed Solution for Test: Bohr’s Model of Atom (April 21) - Question 4


Also, angular momentum is quantised








Test: Bohr’s Model of Atom (April 21) - Question 5

Energy of the electron in nth orbit is given by E   Wavelength of light required to excite an electron in an H-atom from level n = 1 to n = 2 will be (h = 6.62 x 10-34 J s ; c = 3.0 x 108ms -1)

[AIEEE 2012]

Detailed Solution for Test: Bohr’s Model of Atom (April 21) - Question 5

For H - atom , Z = 1




Test: Bohr’s Model of Atom (April 21) - Question 6

If the radius of the first Bohr orbit is x, then de-Broglie wavelength of the electron in the third orbit is nearly

Detailed Solution for Test: Bohr’s Model of Atom (April 21) - Question 6

Angular momentum is quantised , hence
 


Test: Bohr’s Model of Atom (April 21) - Question 7

Direction (Q. Nos. 12-13) This section contains a paragraph, wach describing  theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given  ptions  (a),(b),(c),(d)

Radius of Bohr’s orbit of H-atom is 52.9 pm. An emission in H-atom starts from the orbit having radius 1.3225 nm and ends at 211.6 pm.

Q. 

Wavelength (in nm) associated with this emission is

Detailed Solution for Test: Bohr’s Model of Atom (April 21) - Question 7






Test: Bohr’s Model of Atom (April 21) - Question 8

Radius of Bohr’s orbit of H-atom is 52.9 pm. An emission in H-atom starts from the orbit having radius 1.3225 nm and ends at 211.6 pm.

Q.  Spectral line appears in .......... region.

Detailed Solution for Test: Bohr’s Model of Atom (April 21) - Question 8

Emission (n2 = 5 to n1 = 2) is called Balmer series and appears in visible region.

Test: Bohr’s Model of Atom (April 21) - Question 9

lf En = total energy, Kn = kinetic energy, Vn = potential energy and rn = radius of the nth orbit, then based on Bohr’s theory, match the parameter in Column I with the values in Column II.

Detailed Solution for Test: Bohr’s Model of Atom (April 21) - Question 9





*Answer can only contain numeric values
Test: Bohr’s Model of Atom (April 21) - Question 10

Find the number of waves made by a Bohr’s electron in one complete revolution in its 3rd orbit


Detailed Solution for Test: Bohr’s Model of Atom (April 21) - Question 10

For radius r,
circumference of the orbit = 2πrn


Number of waves in one complete revolution = nλ
For third orbit = nλ = 3λ
Thus, three waves are formed in one revolution. 

360 tests
Information about Test: Bohr’s Model of Atom (April 21) Page
In this test you can find the Exam questions for Test: Bohr’s Model of Atom (April 21) solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Bohr’s Model of Atom (April 21), EduRev gives you an ample number of Online tests for practice

Top Courses for JEE

Download as PDF

Top Courses for JEE