12.10 In accordance with the Bohr’s model, find the quantum number that characterizes the earth’s revolution around the sun in an orbit of radius 1.5 × <math> <mi> </mi> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>11</mn> </mrow> </mrow> </msup> </math> m with orbital speed 3 × <math> <mo>×</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msup> </math> m/s. (Mass of earth = 6.0 × <math> <mo>×</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>24</mn> </mrow> </mrow> </msup> </math> (kg.)

12.10 Radius of the Earth's orbit around the Sun, r = 1.5 * 10 11 mOrbital speed of Earth, v = 3 * 10 4 m/sMass of the Earth, m = 6 * 10 24 kgAccording to Bohr's model, angular momentum is given as: m v r = n h 2 ? , whereh = Planck's constant = 6.626 * 10 - 34 Jsn = Quantum numberHence, n= 2 ? m v r h = 2 * ? * 6 * 10 24 * 3 * 10 4 * 1.5 * 10 11 6.626 * 10 - 34 = 2.56 * 10 74 Hence, the quantum number that characterizes earth's revolution is 2.6 * 10 74

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Physics Ncert Solutions Class 12th Physics Class 12th Atoms

12.10 In accordance with the Bohr’s model, find the quantum number that characterizes the earth’s revolution around the sun in an orbit of radius 1.5 × $\times 10^{11}$ m with orbital speed 3 × $\times 10^{4}$ m/s. (Mass of earth = 6.0 × $\times 10^{24}$ (kg.)

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Vishal Baghel | Contributor-Level 10

5 months ago

12.10 Radius of the Earth's orbit around the Sun, r = 1.5 $* 10^{11}$ m

Orbital speed of Earth, v = 3 $* 10^{4}$ m/s

Mass of the Earth, m = 6 $* 10^{24}$ kg

According to Bohr's model, angular momentum is given as:

$m v r$ = $\frac{n h}{2 ?}$ , where

h = Planck's constant = 6.626 $* 10^{- 34}$ Js

n = Quantum number

Hence, n= $\frac{2 ? m v r}{h}$ = $\frac{2 * ? * 6 * 10^{24} * 3 * 10^{4} * 1.5 * 10^{11}}{6.626 * 10^{- 34}}$ = 2.56 $* 10^{74}$

Hence, the quantum number that characterizes earth's revolution is 2.6 $* 10^{74}$

12.10 Radius of the Earth's orbit around the Sun, r = 1.5 <math> <mi> </mi> <mo>*</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>11</mn> </mrow> </mrow> </msup> </math> mOrbital speed of Earth, v = 3 <math> <mo>*</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msup> </math> m/sMass of the Earth, m = 6 <math> <mo>*</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>24</mn> </mrow> </mrow> </msup> </math> kgAccording to Bohr's model, angular momentum is given as: <math> <mi>m</mi> <mi>v</mi> <mi>r</mi> <mi> </mi> </math> = <math> <mfrac> <mrow> <mrow> <mi>n</mi> <mi>h</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi>? </mi> </mrow> </mrow> </mfrac> </math> , whereh = Planck's constant = 6.626 <math> <mo>*</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>34</mn> </mrow> </mrow> </msup> </math> Jsn = Quantum numberHence, n= <math> <mfrac> <mrow> <mrow> <mn>2</mn> <mi>? </mi> <mi>m</mi> <mi>v</mi> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <mn>2</mn> <mo>*</mo> <mi>? </mi> <mo>*</mo> <mn>6</mn> <mi mathvariant="normal"> </mi> <mo>*</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>24</mn> </mrow> </mrow> </msup> <mo>*</mo> <mn>3</mn> <mi mathvariant="normal"> </mi> <mo>*</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msup> <mo>*</mo> <mn>1.5</mn> <mi> </mi> <mo>*</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>11</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>6.626</mn> <mi mathvariant="normal"> </mi> <mo>*</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>34</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> = 2.56 <math> <mo>*</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>74</mn> </mrow> </mrow> </msup> </math> Hence, the quantum number that characterizes earth's revolution is 2.6 <math> <mo>*</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>74</mn> </mrow> </mrow> </msup> </math>

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12.10 In accordance with the Bohr’s model, find the quantum number that characterizes the earth’s revolution around the sun in an orbit of radius 1.5 × $\times 10^{11}$ m with orbital speed 3 × $\times 10^{4}$ m/s. (Mass of earth = 6.0 × $\times 10^{24}$ (kg.)

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12.10 In accordance with the Bohr’s model, find the quantum number that characterizes the earth’s revolution around the sun in an orbit of radius 1.5 × × 10 11 m with orbital speed 3 × × 10 4 m/s. (Mass of earth = 6.0 × × 10 24 (kg.)

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12.10 In accordance with the Bohr’s model, find the quantum number that characterizes the earth’s revolution around the sun in an orbit of radius 1.5 × $\times 10^{11}$ m with orbital speed 3 × $\times 10^{4}$ m/s. (Mass of earth = 6.0 × $\times 10^{24}$ (kg.)