Question

12.10 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 x 10" m with orbital speed 3 x 10^ m/s. (Mass of earth
= 6.0 x 1024 kg.)​

Answer 1

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r = 1.5×10¹¹m; v = 3×10⁴m/s; M is the mass of earth.

According to Bohr model, M v r = nh/2π.

Thus, n=2.6×10(small74) is the quanta number that charaactrizes the earth revolution.

Answer 2

Answer:

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

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 = nh/2π

Where,

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

n = Quantum number

∴ n = mvr2π/h

= (2πx6x$10^{24}$x3x$10^{4}$x1.5x$10^{11}$)/(6.62x$10^{-34}$ )

= 25.61x1073 = 2.6 x $10^{74}$  

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