PYQSide ExamsMy practice

Atoms · Question 7

JEE Main 2026 (Online) 23rd January Morning ShiftMCQEasy📈 Topic in 16% papers+4−1
1

In hydrogen atom spectrum, (R Rydberg's constant) Choose the correct answer from the options given below :

  1. A, B Only
  2. B, D Only
  3. A, B and D Only
  4. A, B and C Only

Answer

(4) A, B and C Only

Solution

For hydrogen, 1 = R ( 1n12 - 1n22 )

A: Lyman (n1 = 1), maximum wavelength for n2 = 2: 1 = R (1 - 14 ) = 3R4 max = 43R. Correct.

B: As per NCERT, the Balmer series lines lie in the visible region of the spectrum. Correct.

C: Paschen (n1 = 3), minimum wavelength for n2 = : 1 = R9 min = 9R. Correct.

D: Lyman, minimum wavelength for n2 = : 1 = R min = 1R, not 54R. Incorrect.

A, B and C are correct.

Answer: (4)

JEE Main 2026 (Online) 23rd January Morning Shift

Keep practising — Atoms

2

A 7.9 MeV -particle scatters from a target material of atomic number 79. From the given data the estimated diameter of nuclei of the target material is (approximately) m. [ 14 0 = 9 109 N m2/C2 . and electron charge . = 1.6 10-19 C ]

  1. 2.88 10-14
  2. 5.76 10-14
  3. 1.44 10-13
  4. 1.69 10-12
JEE Main 2026 (Online) 22nd January Morning Shift
3

The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly nm.

  1. 1784
  2. 1875
  3. 1217
  4. 1550
JEE Main 2026 (Online) 22nd January Evening Shift
4

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is m. (Atomic number of gold = 79 and 14 0 = 9 109 in SI units)

  1. 2.95 10-16
  2. 3.85 10-14
  3. 2.95 10-14
  4. 3.85 10-16
JEE Main 2026 (Online) 21st January Morning Shift
5

The energy of an electron in an orbit of the Bohr's atom is -0.04E0 eV where E0 is the ground state energy. If L is the angular momentum of the electron in this orbit and h is the Planck's constant, then 2 Lh is :

  1. 6
  2. 2
  3. 5
  4. 4
JEE Main 2026 (Online) 21st January Evening Shift
6

In a hydrogen like ion, the energy difference between the 2nd excitation energy state and ground is 108.8 eV. The atomic number of the ion is:

  1. 1
  2. 4
  3. 3
  4. 2
JEE Main 2025 (Online) 7th April Morning Shift
7

For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is

  1. 5 : 27
  2. 27 : 5
  3. 3 : 4
  4. 5 : 36
JEE Main 2025 (Online) 7th April Morning Shift
8

Considering the Bohr model of hydrogen-like atoms, the ratio of the radius of 5th orbit of the electron in Li2+ and He+ is

  1. 32
  2. 23
  3. 49
  4. 94
JEE Main 2025 (Online) 4th April Morning Shift
9

Given below are two statements : In the light of the above statements, choose the most appropriate answer from the options given below :

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Both Statement I and Statement II are incorrect
  4. Statement I is incorrect but Statement II is correct
JEE Main 2025 (Online) 4th April Evening Shift
10

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. In the light of the above statements, choose the correct answer from the options given below

  1. A is true but R is false
  2. Both A and R are true and R is the correct explanation of A
  3. A is false but R is true
  4. Both A and R are true but R is NOT the correct explanation of A
JEE Main 2025 (Online) 3rd April Evening Shift
11

An electron in the hydrogen atom initially in the fourth excited state makes a transition to nth energy state by emitting a photon of energy 2.86 eV. The integer value of n will be .

JEE Main 2025 (Online) 3rd April Evening Shift

All Atoms PYQs →

PDFDownload Atoms PYQs as PDF Free, with answer key and a QR code to every solution — print it or study offline⬇
All JEE Main Physics chapters