The Electron's Speed in Hydrogen

Q. It is found experimentally that 13.6 eV of energy is required to separate a hydrogen atom into a proton and an electron. Taking the electron's mass as 9.1 × 10⁻³¹ kg, What is the speed of the electron in its orbit, in m/s? expects 2.2 × 10⁶ m/s

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Q. It is found experimentally that 13.6 eV $E_{b} = 13.6$ eV of energy is required to separate a hydrogen atom into a proton and an electron. Taking the electron's mass as 9.1 × 10⁻³¹ kg $m = 9.1 \times 10^{−31}$ kg, What is the speed of the electron in its orbit, in m/s?

The playground

Inputs

energy

E_{b}

13.6 eV

mass

m

9.1e-31 kg

Outputs

speed

v

The solution

Solved with BohrModel

Given

$E_{b} = 13.6$ eV
$m = 9.1 \times 10^{−31}$ kg

Find $v$ .

Working

$E_{b} = 13.6$ eV $= 2.17896022 \times 10^{−18}$ J
energy to free the electron(electron) $E_{b} = \frac{k \cdot e^{2}}{2 \cdot r}$
$2.17896022 \times 10^{−18} = \frac{8.98755179 \times 10^{9} \cdot {(1.60217663 \times 10^{−19})}^{2}}{2 \cdot r}$
$r = 5.29398731 \times 10^{−11}$ m
held in orbit by the proton(electron) $\frac{m \cdot v^{2}}{r} = \frac{k \cdot e^{2}}{r^{2}}$
$\frac{9.1 \times 10^{−31} \cdot v^{2}}{5.29398731 \times 10^{−11}} = \frac{8.98755179 \times 10^{9} \cdot {(1.60217663 \times 10^{−19})}^{2}}{{(5.29398731 \times 10^{−11})}^{2}}$
$v = 2.188 \times 10^{6}$ m/s
$v \approx 2.2 \times 10^{6}$ m/sto 2 significant figures

Conclusion

The proton's pull is what keeps the electron circling, so the orbit's size fixes how fast it must go.

The Ruby

1require_relative "../physics"23question :the_electron_s_speed_in_hydrogen do4  kind :numerical5  describes "It is found experimentally that %{energy} of energy is required to separate a hydrogen atom into a " \6            "proton and an electron. Taking the electron's mass as %{mass}, %{speed}, in m/s?"7  given energy: "13.6 eV"8  given mass: "9.1e-31 kg"9  asks speed: "What is the speed of the electron in its orbit"10end

1require_relative "../questions/the_electron_s_speed_in_hydrogen"2require_relative "../science/atoms/bohr_model"34solve :the_electron_s_speed_in_hydrogen, using: BohrModel do5  objects :electron67  take energy: electron.binding_energy8  take mass: electron.mass910  speed = electron.orbital_speed11  answer speed:, in: "m/s", figures: 21213  conclusion "The proton's pull is what keeps the electron circling, so the orbit's size fixes how fast it must go."14end

1require_relative "../solutions/the_electron_s_speed_in_hydrogen"23play_with :the_electron_s_speed_in_hydrogen do4  title "The Electron's Speed in Hydrogen"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 electron	$E_{b}$	J	—
orbit radius of electron	$r$	m	—
mass of ground	$m_{g}$	kg	0 … ∞
orbital speed of ground	$v_{g}$	m/s	0 … ∞
mass of electron	$m$	kg	0 … ∞
orbital speed of electron	$v$	m/s	0 … ∞
atomic number of ground	$Z_{g}$	—	—
quantum number of ground	$n_{g}$	—	—
atomic number of electron	$Z$	—	—
quantum number of electron	$n$	—	—
potential energy of ground	$U_{g}$	J	—
potential energy of electron	$U$	J	—
total energy of ground	$E_{g}$	J	—
total energy of electron	$E$	J	—
photon energy of ground	$E_{p,g}$	J	—
frequency of ground	$ν_{g}$	Hz	0 … ∞
photon energy of electron	$E_{p}$	J	—
frequency of electron	$ν$	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(electron)

\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(electron)

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(electron)

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(electron)

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(electron)

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(electron)

E_{p,g} = h \cdot ν_{g}

one photon for one jump(ground)

E_{p} = h \cdot ν

one photon for one jump(electron)