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Atoms · Question 5

JEE Main 2026 (Online) 2nd April Evening ShiftNumericalMedium📈 Topic in 35% papers+4
1

Using Bohr's model, calculate the ratio of the magnetic fields generated due to the motion of the electrons in the 2nd and 4th orbits of hydrogen atom .

Answer

32

Solution

The orbiting electron forms a current loop: I = eT = ev2 r

Field at the nucleus: B = 0I2r = 0ev4 r2 ⇒ B vr2

Bohr's model (Z = 1): v 1n, r n2

⇒ B 1/nn4 = 1n5

B2B4 = ( 42 )5 = 25 = 32

Answer: 32

JEE Main 2026 (Online) 2nd April Evening Shift

Keep practising — Atoms

2

Two electrons are moving in orbits of two hydrogen-like atoms with speeds 3 105 m/s and 2.5 105 m/s respectively. If the radii of these orbits are nearly same then the possible order of energy states are respectively.

  1. 8 and 10
  2. 10 and 12
  3. 9 and 8
  4. 6 and 5
JEE Main 2026 (Online) 24th January Morning Shift
3

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
JEE Main 2026 (Online) 23rd January Morning Shift
4

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
5

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
6

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
7

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
8

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
9

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
10

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
11

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

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