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 1.5×10^11 m

Answer 1

domalel4

gomole of He-

Total of moles

L

=1

1 Solution has 46%, wher) GHOH in weater

and molo prication of CHE OH.

Super bapPm en solution

yer de present in 16 of salon

Answer 2

Answer:

2.6*$10^{74}$

Step-by-step explanation:

 Radius of the orbit of the Earth around the Sun, r = 1.5 × $10^{11}$ m

Orbital speed of the Earth, ν = 3 × $10^{4}$ m/s

Mass of the Earth, m = 6.0 × $10^{24}$ kg

According to Bohr’s model, angular momentum is quantized and given as: mvr=n h/2π ---------eqn 1

Where, h = Planck’s constant = 6.62 × $10^{-34}$ Js

n = Quantum number

∴from eqn. 1 n=mvr2π/h.

Putting all the values in the above equation we get

                  ⇒  n= 2.6×$10^{74}$

Hence, the quanta number that characterizes the Earth’ revolution is 2.6 × $10^{74}$ .