Calculate the velocity of electron in bohr’s first orbit of hydrogen atom .How many times does the electron go in Bohr’s first orbit in one second
The correct answer and explanation is:
Correct Answer:
Velocity of the electron in Bohr’s first orbit:
v=2.18×106 m/sv = 2.18 \times 10^6 \, \text{m/s}
Number of revolutions per second (frequency):
f=6.58×1015 revolutions/secondf = 6.58 \times 10^{15} \, \text{revolutions/second}
Explanation:
In Bohr’s model of the hydrogen atom, the electron revolves around the nucleus in certain allowed circular orbits without radiating energy. The velocity of the electron in the first orbit (n = 1) is determined by using Bohr’s formula: vn=2.18×106 m/snv_n = \frac{2.18 \times 10^6 \, \text{m/s}}{n}
For the first orbit (n = 1): v1=2.18×106 m/sv_1 = 2.18 \times 10^6 \, \text{m/s}
This is the speed of the electron in meters per second in Bohr’s first orbit.
To find how many times the electron goes around the nucleus in one second, we need the circumference of the orbit and then divide the speed by this distance to get the frequency of revolution.
The radius of the first Bohr orbit is: r1=5.29×10−11 mr_1 = 5.29 \times 10^{-11} \, \text{m}
The circumference of the orbit is: C=2πr=2π×5.29×10−11=3.32×10−10 mC = 2\pi r = 2\pi \times 5.29 \times 10^{-11} = 3.32 \times 10^{-10} \, \text{m}
Now, frequency of revolution is: f=vC=2.18×1063.32×10−10≈6.58×1015 revolutions/secondf = \frac{v}{C} = \frac{2.18 \times 10^6}{3.32 \times 10^{-10}} \approx 6.58 \times 10^{15} \, \text{revolutions/second}
This shows the electron completes over six quadrillion revolutions per second in Bohr’s first orbit. This extremely high frequency reflects the tiny scale and high energy of atomic-level systems. Bohr’s model helped lay the foundation for quantum mechanics by introducing the idea of quantized orbits, which successfully explained the spectral lines of hydrogen.