Question

In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 1011 m with orbital speed 3 × 104 m/s. (Mass of earth = 6.0 × 1024 kg.)

Answer 1

Hey Dear,

◆ Answer -

n = 2.559×10^74

● Explaination -

# Given -

m = 6×10^24 kg

v = 3×10^4 m/s

r = 1.5×10^11 m

# Solution -

In accordance with the Bohr's model,

mvr = nh/2π

Rearranging it, we will get -

n = 2πmvr/h

n = 2 × 3.142 × 6×10^24 × 3×10^4 × 1.5×10^11 / 6.63×10^-34

n = 2.559×10^74

Therefore, quantum number associated with earth's revolution is 2.559×10^74 .

Thanks dear. Hope that helps you..

Answer 2

Answer:

n =$2.6 \times {10}^{74}$