JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Atomic Structure

WhatsApp Email

Frequency of the de-Broglie wave of electron in Bohr's first orbit of hydrogen atom is ______ × 10¹³ Hz (nearest integer).
[Given: Rₕ (Rydberg constant) = 2.18 × 10⁻¹⁸ J. h (Plank's constant) = 6.6 × 10⁻³⁴ J.s.]

Answer: 661

Solution

The de-Broglie wave frequency is

ν = \frac{v}{λ}

, and since

λ = \frac{h}{m v}

this becomes

ν = \frac{m v ^{2}}{h} = \frac{2 ( \frac{1}{2} m v ^{2} )}{h}

. In the first Bohr orbit the kinetic energy of the electron equals the magnitude of its ground-state energy,

\frac{1}{2} m v^{2} = 2.18 \times 1 0^{- 18}

J. Substituting,

ν = \frac{2 \times 2.18 \times 1 0 ^{- 18}}{6.6 \times 1 0 ^{- 34}} = 6.606 \times 1 0^{15}

= 660.6 \times 1 0^{13}

Hz, which rounds to

661 \times 1 0^{13}

Hz.

JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Integer answer

Practise more: all JEE Main PYQs on Atomic Structure →

More JEE Main questions on Atomic Structure

Given below are two statements:

Statement I: When an electric discharge is passed through gaseous hydrogen, the hydrogen molecules dissociate and the energetically excited hydrogen atoms produce electromagnetic radiation of discrete frequencies.

Statement II: The frequency of s …
JEE Main 2026View solution →
The energy required by electrons, present in the first Bohr orbit of hydrogen atom to be excited to second Bohr orbit is ____ $J mol^{- 1}$ .

Given : $R_{H} = 2.18 \times 1 0^{- 11} ergs$ .
JEE Main 2026View solution →
Consider the following spectral lines for atomic hydrogen:
A. First line of Paschen series
B. Second line of Balmer series
C. Third line of Paschen series
D. Fourth line of Bracket series

The correct arrangement of the above lines in ascending order of energy is:
JEE Main 2026View solution →
The energy of first (lowest) Balmer line of H atom is x J. The energy (in J) of second Balmer line of H atom is :
JEE Main 2026View solution →
Which of the following statements regarding the energy of the stationary state is true in the following one-electron system ?
JEE Main 2026View solution →
Given,
(A) $n = 5, m_{1} = - 1$
(B) $n = 3, 1 = 2, m_{1} = - 1, m_{s} = + \frac{1}{2}$

The maximum number of electron(s) in an atom that can have the quantum numbers as given in (A) and (B) respectively are :
JEE Main 2026View solution →