A 10kg satellite circles earth once every 2 hour in an orbit having a radius 8000km . assuming that bohr's angular momentum postulate applies to satellite just as it does to an electron in the hydrogen atom find the quantum number of the orbit of the satellite
Answers
The quantum number of the orbit of the satellite is 5.3 * 10⁴⁵.
Explanation:
=> It is given that,
mass of satellite, m = 10 kg
Period, T = 2 hour = 2 * 3600 sec = 7200 sec
Radius of an orbit, R = 8000 km = 8000 * 10³ m = 8 * 10⁶ m
=> Angular frequency, ω:
ω = 2π/T
=> As per the Bohr's postulate of quantized angular momentum:
mvR = nh/2π
But, v = R*ω
∴ m(R*ω)R = nh/2π
mωR² = nh/2π
m(2π/ T)R² = nh/2π [∵ ω = 2π / T]
n = 4π²mR² / T*h
=> By placing the given values in above equation, we get
[n = 4π²mR² / T*h]
n = 4 * 3.14 * 3.14 * 10 * 8 * 10⁶ * 8 * 10⁶ / 7200 * 6.63x 10⁻³⁴
= 0.0528 * 10⁴⁷
≈ 5.3 * 10⁴⁵
Thus, the quantum number of the orbit of the satellite is 5.3 * 10⁴⁵.
Learn more:
Q:1 A satellite revolves around the earth in a circular orbit once 90 minutes. If the radius of earth is 6400 km, how high is the satellite above the earth? (g= 9.8 metre per second square)
Click here: https://brainly.in/question/1629716
Q:2 Find the principle quantum number for the electron having the de broglie wavelength in an orbit of hydrogen atom is 10^-9 m.
Click here: https://brainly.in/question/8077279
Answer:
5.3×10^45
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