Velocity of Electron in First Bohr Orbit | JEE Mains PYQ | Structure of Atom
Q. The value of Rydberg constant (RH) is 2.18 × 10−18 J. The velocity of electron having mass 9.1 × 10−31 kg in Bohr's first orbit of hydrogen atom = ______ × 105 m s−1 (nearest integer).
Solution
Step 1: Energy of electron in first Bohr orbit

For hydrogen atom,
Energy in first orbit:
E = −RH

Step 2: Relation between kinetic energy and total energy

For Bohr orbit:
Total energy E = − K.E.

So,
K.E. = RH = 2.18 × 10−18 J

Step 3: Expression for kinetic energy

K.E. = 1/2 m v²

Step 4: Substitute values

1/2 × 9.1 × 10−31 × v² = 2.18 × 10−18

v² =
(2 × 2.18 × 10−18) / (9.1 × 10−31)

v² ≈ 4.79 × 1012

v ≈ 2.19 × 106 m s−1

Step 5: Express in required format

v = 21.9 × 105 m s−1

Final Answer = 22

Previous Year Questions – Structure of Atom
In an atom, total number of electrons having quantum numbers n = 4, ml = 1 and ms = −1/2 is ______.
In an atom, total number of electrons having quantum numbers n = 4, ml = 1 and ms = −1/2 is ______.
For hydrogen atom, energy of an electron in first excited state is −3.4 eV. K.E. of the same electron of hydrogen atom is x eV. Value of x is ______ × 10⁻¹ eV.
Frequency of the de-Broglie wave of electron in Bohr's first orbit of hydrogen atom is ______ × 10¹³ Hz.
Given below are two statements related to Heisenberg uncertainty principle. Choose the correct answer.
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