The Ground State's Potential Energy

Q. The ground state energy of hydrogen atom is -13.6 eV. What is the potential energy of the electron in this state, in eV? expects -27.2 eV

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Q. The ground state energy of hydrogen atom is -13.6 eV $E = -13.6$ eV. What is the potential energy of the electron in this state, in eV?

The playground

Inputs

energy

E

-13.6 eV

Outputs

potential

U

The solution

Solved with BohrModel

Given

$E = -13.6$ eV

Find $U$ .

Working

$E = -13.6$ eV $= −2.17896022 \times 10^{−18}$ J
energy of the atom(electron) $E = \frac{- k \cdot e^{2}}{2 \cdot r}$
$(−2.17896022 \times 10^{−18}) = \frac{- 8.98755179 \times 10^{9} \cdot {(1.60217663 \times 10^{−19})}^{2}}{2 \cdot r}$
$r = 5.29398731 \times 10^{−11}$ m
energy of its place(electron) $U = \frac{- k \cdot e^{2}}{r}$
$U = \frac{- 8.98755179 \times 10^{9} \cdot {(1.60217663 \times 10^{−19})}^{2}}{5.29398731 \times 10^{−11}}$
$U = −27.2$ eV
$U = −4.35792044 \times 10^{−18}$ J $= −27.2$ eV
$U \approx −27.2$ eVto 3 significant figures

Conclusion

Held in a circle by the proton's pull, the electron's potential energy is twice as deep as the atom's energy, half of it given back as the energy of its motion.

The Ruby

1require_relative "../physics"23question :the_ground_state_s_potential_energy do4  kind :numerical5  describes "The ground state energy of hydrogen atom is %{energy}. %{potential} of the electron in this state, in eV?"6  given energy: "-13.6 eV"7  asks potential: "What is the potential energy"8end

1require_relative "../questions/the_ground_state_s_potential_energy"2require_relative "../science/atoms/bohr_model"34solve :the_ground_state_s_potential_energy, using: BohrModel do5  objects :electron67  take energy: electron.total_energy89  potential = electron.potential_energy10  answer potential:, in: "eV", figures: 31112  conclusion "Held in a circle by the proton's pull, the electron's potential energy is twice as deep as the " \13             "atom's energy, half of it given back as the energy of its motion."14end

1require_relative "../solutions/the_ground_state_s_potential_energy"23play_with :the_ground_state_s_potential_energy do4  title "The Ground State's Potential Energy"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)