The Earth's Quantum Number

Q. In accordance with the Bohr's model, what is 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.) expects 2.6 × 10⁷⁴

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Q. In accordance with the Bohr's model, what is the quantum number that characterises the earth's revolution around the sun in an orbit of radius 1.5 × 10¹¹ m $r = 1.5 \times 10^{11}$ m with orbital speed 3 × 10⁴ m/s $v = 3 \times 10^{4}$ m/s? (Mass of earth = 6.0 × 10²⁴ kg $m = 6.0 \times 10^{24}$ kg.)

The playground

Inputs

radius

r

150000000000.00 m

speed

v

30000.0 m/s

mass

m

6e+24 kg

Outputs

number

n

The solution

Solved with BohrModel

Given

$r = 1.5 \times 10^{11}$ m
$v = 3 \times 10^{4}$ m/s
$m = 6.0 \times 10^{24}$ kg

Find $n$ .

Working

angular momentum comes in steps(earth) $m \cdot v \cdot r = \frac{n \cdot h}{2 \cdot π}$
$6 \times 10^{24} \cdot 30000 \cdot 1.5 \times 10^{11} = \frac{n \cdot 6.62607015 \times 10^{−34}}{2 \cdot 3.14159265}$
$n = 2.56028082 \times 10^{74}$
$n \approx 2.6 \times 10^{74}$ to 2 significant figures

Conclusion

Angular momentum comes in steps of Planck's constant over two pi, but the earth's is so vast that the steps are far too fine to notice: quantum rules fade into classical ones at large sizes.

The Ruby

1require_relative "../physics"23question :the_earth_s_quantum_number do4  kind :numerical5  describes "In accordance with the Bohr's model, %{number} that characterises the earth's revolution around the " \6            "sun in an orbit of radius %{radius} with orbital speed %{speed}? (Mass of earth = %{mass}.)"7  given radius: "1.5e11 m"8  given speed: "3e4 m/s"9  given mass: "6.0e24 kg"10  asks number: "what is the quantum number"11end

1require_relative "../questions/the_earth_s_quantum_number"2require_relative "../science/atoms/bohr_model"34solve :the_earth_s_quantum_number, using: BohrModel do5  objects :earth67  take radius: earth.orbit_radius8  take speed: earth.orbital_speed9  take mass: earth.mass1011  number = earth.quantum_number12  answer number:, figures: 21314  conclusion "Angular momentum comes in steps of Planck's constant over two pi, but the earth's is so vast that the " \15             "steps are far too fine to notice: quantum rules fade into classical ones at large sizes."16end

1require_relative "../solutions/the_earth_s_quantum_number"23play_with :the_earth_s_quantum_number do4  title "The Earth's Quantum Number"5end

The laws

quantity	written	units	within	from
pi	$π$	—	0 … ∞	BohrModel
coulomb constant	$k$	N m² C⁻²	—	BohrModel
elementary charge	$e$	C	—	BohrModel
plancks constant	$h$	J s	—	BohrModel
mass	$m$	kg	0 … ∞	BohrModel
orbit radius	$r$	m	—	BohrModel
orbital speed	$v$	m/s	0 … ∞	BohrModel
binding energy	$E_{b}$	J	—	BohrModel
quantum number	$n$	—	—	BohrModel
atomic number	$Z$	—	—	BohrModel
potential energy	$U$	J	—	BohrModel
total energy	$E$	J	—	BohrModel
photon energy	$E_{p}$	J	—	BohrModel
frequency	$ν$	Hz	0 … ∞	BohrModel
binding energy of ground	$E_{b,g}$	J	—
orbit radius of ground	$r_{g}$	m	—
binding energy of earth	$E_{b}$	J	—
orbit radius of earth	$r$	m	—
mass of ground	$m_{g}$	kg	0 … ∞
orbital speed of ground	$v_{g}$	m/s	0 … ∞
mass of earth	$m$	kg	0 … ∞
orbital speed of earth	$v$	m/s	0 … ∞
atomic number of ground	$Z_{g}$	—	—
quantum number of ground	$n_{g}$	—	—
atomic number of earth	$Z$	—	—
quantum number of earth	$n$	—	—
potential energy of ground	$U_{g}$	J	—
potential energy of earth	$U$	J	—
total energy of ground	$E_{g}$	J	—
total energy of earth	$E$	J	—
photon energy of ground	$E_{p,g}$	J	—
frequency of ground	$ν_{g}$	Hz	0 … ∞
photon energy of earth	$E_{p}$	J	—
frequency of earth	$ν$	Hz	0 … ∞

E_{b,e} = \frac{k \cdot e^{2}}{2 \cdot r_{e}}

energy to free the electron

\frac{m_{e} \cdot {v_{e}}^{2}}{r_{e}} = \frac{k \cdot e^{2}}{{r_{e}}^{2}}

held in orbit by the proton

v_{e} = \frac{2 \cdot π \cdot k \cdot Z_{e} \cdot e^{2}}{n_{e} \cdot h}

speed set by the level

m_{e} \cdot v_{e} \cdot r_{e} = \frac{n_{e} \cdot h}{2 \cdot π}

angular momentum comes in steps

U_{e} = \frac{- k \cdot e^{2}}{r_{e}}

energy of its place

E_{e} = \frac{- k \cdot e^{2}}{2 \cdot r_{e}}

energy of the atom

E_{p,l} = h \cdot ν_{l}

one photon for one jump

E_{b,g} = \frac{k \cdot e^{2}}{2 \cdot r_{g}}

energy to free the electron(ground)

E_{b} = \frac{k \cdot e^{2}}{2 \cdot r}

energy to free the electron(earth)

\frac{m_{g} \cdot {v_{g}}^{2}}{r_{g}} = \frac{k \cdot e^{2}}{{r_{g}}^{2}}

held in orbit by the proton(ground)

\frac{m \cdot v^{2}}{r} = \frac{k \cdot e^{2}}{r^{2}}

held in orbit by the proton(earth)

v_{g} = \frac{2 \cdot π \cdot k \cdot Z_{g} \cdot e^{2}}{n_{g} \cdot h}

speed set by the level(ground)

v = \frac{2 \cdot π \cdot k \cdot Z \cdot e^{2}}{n \cdot h}

speed set by the level(earth)

m_{g} \cdot v_{g} \cdot r_{g} = \frac{n_{g} \cdot h}{2 \cdot π}

angular momentum comes in steps(ground)

m \cdot v \cdot r = \frac{n \cdot h}{2 \cdot π}

angular momentum comes in steps(earth)

U_{g} = \frac{- k \cdot e^{2}}{r_{g}}

energy of its place(ground)

U = \frac{- k \cdot e^{2}}{r}

energy of its place(earth)

E_{g} = \frac{- k \cdot e^{2}}{2 \cdot r_{g}}

energy of the atom(ground)

E = \frac{- k \cdot e^{2}}{2 \cdot r}

energy of the atom(earth)

E_{p,g} = h \cdot ν_{g}

one photon for one jump(ground)

E_{p} = h \cdot ν

one photon for one jump(earth)