The value of Rydberg constant (R_H) 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 ms⁻¹ (nearest integer).

Solution

Chemistry ma’am · 3 steps

Kinetic energy in a Bohr orbit equals minus the total energy, so set 1/2mv² = R_H.

1. Energy link. In any Bohr orbit, kinetic energy = - total energy. For n = 1 in hydrogen the total energy is -R_H = -2.18 × 10⁻¹⁸ J, so the kinetic energy is +2.18 × 10⁻¹⁸ J.

   1/2mv² = R_H ⇒ v = √(2R_H/m)

2. Put in numbers. v = √((2 × 2.18 × 10⁻¹⁸)/(9.1 × 10⁻³¹)) = √(4.79 × 10¹²) = 2.19 × 10⁶ m s⁻¹.
3. So. v = 21.9 × 10⁵ m s⁻¹, which rounds to 22.

The common slip: 15. Leaving out the factor of 2 and writing mv² = R_H gives 1.55 × 10⁶ m s⁻¹, that is 15, not 22.

The idea to keep. For a Bohr orbit, kinetic energy is equal and opposite to the total energy. Once you know the orbit energy, 1/2mv² gives the speed directly.

Answer from the NTA final answer key. Solution by PrepWiser.