Consider aiming a beam of free electrons towards free protons. When they scatter, an electron and a proton can

NCERT Exemplar · Q25

Q. Consider aiming a beam of free electrons towards free protons. When they scatter, an electron and a proton cannot combine to produce a H-atom,

(a) because of energy conservation.

(b) without simultaneously releasing energy in the form of radiation.

(d) because of angular momentum conservation.

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Concept understanding — Photon Energy

Photon Energy: The Energy Carried by Light

Light does not carry its energy in a continuous stream. It comes in discrete packets — like individual drops instead of a steady hose. Each packet is called a photon: the smallest possible unit of light of a given frequency. You cannot have half a photon; it is all or nothing.

The Intuition: Colour Determines Energy

Red light and blue light are both light, but they behave differently — blue light causes sunburn and drives chemical reactions more readily than red. The reason is that the colour (frequency) of light directly fixes how much energy each photon carries, and blue photons carry more energy than red photons.

The Planck–Einstein Relation

The energy $E$ of a single photon is directly proportional to its frequency $f$ , and inversely proportional to its wavelength $\lambda$ :

E = h f = \frac{h c}{\lambda}

Where:

$E$ = energy of one photon (joules, J)
$f$ = frequency (hertz, Hz)
$\lambda$ = wavelength (metres, m)
$c$ = speed of light in vacuum ( $3.00 \times 10^8$ m/s)
$h$ = Planck's constant ( $6.626 \times 10^{-34}$ J·s)

Important

Planck's constant $h$ is extremely tiny, so an individual photon carries a very small amount of energy — which is why we never feel single photons striking our skin.

What This Tells You

Higher frequency = higher energy. Gamma rays have extremely energetic photons; radio waves have very low-energy photons.
Shorter wavelength = higher energy. Ultraviolet photons are more energetic than visible-light photons; infrared photons are less.
Energy is quantised. Light energy comes in fixed packets — you can have 1, 2, or 1000 photons, but never 0.5. This was Max Planck's revolutionary idea of 1900.

This is why UV light causes sunburn (UV photons carry enough energy to damage DNA, while visible photons do not), why X-ray photons are energetic enough to pass through soft tissue but get absorbed by bone, and why a photon needs enough energy to actually break a bond or drive a photochemical/biological reaction, rather than merely wavelength-dependent intensity.

A Concrete Example

Note

Question: Which has more energy — a photon of red light ( $\lambda = 700$ nm) or a photon of blue light ( $\lambda = 450$ nm)?

Since $E = \dfrac{hc}{\lambda}$ , energy is inversely proportional to wavelength, so the shorter-wavelength blue photon carries more energy:

Red: $E_{\text{red}} = \dfrac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{700 \times 10^{-9}} \approx 2.84 \times 10^{-19}$ J …

Why this formula?

Why Photon Energy is $E = hf$ — The Reasoning Behind the Formula

The formula $E = hf$ is not something you memorise and plug numbers into. It comes from a deep shift in how physicists understood light.

The problem that forced the formula

By the late 1800s, classical physics said light was a wave — continuous, carrying energy in proportion to its intensity (brightness). But blackbody radiation and the photoelectric effect showed something bizarre: when you shine light on a metal, electrons are ejected only if the light's frequency is above a certain threshold, no matter how bright the light is. Below that frequency, even the brightest lamp won't kick out a single electron. Classically this made no sense — a wave's energy depends on amplitude, not frequency.

Einstein's radical idea (1905)

Einstein proposed that light comes in discrete packets — photons — each carrying a fixed energy that depends only on its frequency, with Planck's constant $h$ as the proportionality:

A single photon's energy is $E = hf$ , where $f$ is the frequency of the light.

This directly explains the photoelectric effect: an electron needs a minimum energy (the work function $\phi$ ) to escape. If a photon's energy $hf < \phi$ , no electron is ejected — regardless of how many photons you send, because each photon interacts with one electron individually.

Important

$E = hf$ is not derived from deeper principles — it is a postulate justified by the experiments it explains, and it unified optics with quantum mechanics.

The equivalent forms

Since $c = f\lambda$ for light in vacuum, the same idea in terms of wavelength is:

E = \frac{hc}{\lambda}

Use this when you're given $\lambda$ instead of $f$ . A photon also carries momentum (despite having no mass): …

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An electron and a proton cannot combine into a hydrogen atom in free space without emitting radiation because both energy and momentum must be conserved simultaneously — a bound state has lower energy, and the excess must go somewhere, but a single photon cannot carry away both the energy difference and the required momentum.

The key insight here is that when a free electron and a free proton come together to form a hydrogen atom, the total energy of the system decreases. The electron goes from having positive kinetic energy (in free space) to being in a bound state with negative total energy. That energy difference — the binding energy — has to go somewhere. It cannot simply vanish.

But there is a deeper constraint. In any scattering or capture process, both energy and momentum must be conserved. If the electron and proton simply combined without emitting anything, the final hydrogen atom would have to carry away all the initial momentum. That is possible in principle, but the problem is that the atom's kinetic energy would then be too large — it would exceed the binding energy released, making the capture impossible without a third particle to carry away the excess.

Let us walk through the reasoning step by step.

Energy conservation alone does not forbid capture. The electron and proton have some initial kinetic energy (say, $K_i$ ). When they bind into a hydrogen atom, the final energy is the atom's ground state energy ( $-13.6\ \text{eV}$ ) plus its kinetic energy as a whole ( $K_f$ ). Energy conservation gives

K_i = (-13.6\ \text{eV}) + K_f + E_{\text{emitted}},

where $E_{\text{emitted}}$ is any energy carried away by radiation. If no radiation is emitted, then $K_f = K_i + 13.6\ \text{eV}$ , which is perfectly allowed by energy conservation alone. So option (A) is false — energy conservation does not prevent capture.

Momentum conservation is the real obstacle.

In the centre-of-mass frame, the total momentum before capture is zero. After capture, if no radiation is emitted, the hydrogen atom must also have zero momentum — so it is at rest. That means its kinetic energy $K_f = 0$ . But then energy conservation would require $K_i = -13.6\ \text{eV}$ , which is impossible since kinetic energy cannot be negative.

In the lab frame, the situation is even clearer: the initial momenta of the electron and proton are generally not equal and opposite. If they combine into a single atom, the atom's momentum must equal the total initial momentum. That is fine — but the atom's kinetic energy is then $p^2/(2M)$ , where $M$ is the atom's mass. This is tiny compared to the binding energy (since $M$ is large), but the binding energy is fixed at $13.6\ \text{eV}$ . The only way to satisfy both conservation laws is to emit a photon that carries away both the excess energy and some momentum.
A single photon cannot do both jobs perfectly.

Suppose the electron and proton capture each other and emit one photon. Energy conservation:

K_i + m_e c^2 + m_p c^2 = E_{\text{atom}} + E_\gamma.

Momentum conservation:

\vec{p}_e + \vec{p}_p = \vec{p}_{\text{atom}} + \vec{p}_\gamma.

The photon's momentum is $E_\gamma / c$ . For a typical capture, the binding energy is $\sim 13.6\ \text{eV}$ , so $E_\gamma \approx 13.6\ \text{eV}$ . The photon's momentum is then $13.6\ \text{eV}/c$ , which is very small. But the initial momenta of the electron and proton can be much larger (even for thermal energies, $k_B T \sim 0.025\ \text{eV}$ gives momenta of order $\sqrt{2mK} \sim \text{a few eV}/c$ ). The atom's recoil momentum must balance the difference, but the atom is heavy — its recoil kinetic energy is $p^2/(2M) \sim ( \text{a few eV}/c)^2 / (2 \times 10^9\ \text{eV}/c^2) \sim 10^{-9}\ \text{eV}$ , negligible. So the photon cannot carry away enough momentum to balance the initial momenta unless the initial kinetic energies are extremely small.

In practice, for free particles at any reasonable temperature, a single photon cannot simultaneously conserve both energy and momentum. This is why free electron-proton capture requires a third body (like another photon or a nearby atom) to take up the slack.

Angular momentum is not the issue. The electron and proton can have any relative angular momentum. The hydrogen atom's ground state has zero orbital angular momentum ( $l=0$ ), but the spins can combine to give either a singlet or triplet state. Angular momentum can be conserved by emitting a photon that carries away one unit of angular momentum (it is a spin-1 particle). So angular momentum conservation does not forbid capture — it just determines the photon's polarisation. Option (D) is false. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
$3$
›Reveal solutionSolution
The key idea is that the 12.75 eV electron excites the hydrogen atom to a specific energy level, and the number of spectral lines equals the number of possible downward transitions from that level. The maximum number of spectral lines is 6.

When a hydrogen atom absorbs energy from a colliding electron, the electron in the atom jumps from its ground state ( $n=1$ ) to a higher energy level. The energy levels of hydrogen are given by $E_n = -\frac{13.6}{n^2}$ eV. The energy difference between the ground state ( $n=1$ ) and any excited state ( $n$ ) is $\Delta E = 13.6 \left(1 - \frac{1}{n^2}\right)$ eV. The bombarding electron provides exactly 12.75 eV of energy, so we need to find which $n$ this corresponds to.

The number of spectral lines emitted when the atom de-excites depends on how many distinct downward jumps are possible. If the atom is excited to level $n$ , the number of spectral lines is the number of ways to go from $n$ down to $1$ , which is $\frac{n(n-1)}{2}$ .

Let’s work through the steps.

Find the excited state reached. The ground state energy is $E_1 = -13.6$ eV. The energy of the $n$ th level is $E_n = -\frac{13.6}{n^2}$ eV. The energy absorbed to go from $n=1$ to $n$ is:

$\Delta E = E_n - E_1 = -\frac{13.6}{n^2} - (-13.6) = 13.6 \left(1 - \frac{1}{n^2}\right) \text{ eV}.$
Set this equal to 12.75 eV:
$13.6 \left(1 - \frac{1}{n^2}\right) = 12.75.$
Divide both sides by 13.6:
$1 - \frac{1}{n^2} = \frac{12.75}{13.6} = 0.9375.$
So:
$\frac{1}{n^2} = 1 - 0.9375 = 0.0625.$
Hence $n^2 = \frac{1}{0.0625} = 16$ , so $n = 4$ .

The atom is excited to the $n=4$ level.

Determine the possible transitions.

From $n=4$ , the electron can fall to any lower level: $n=3$ , $n=2$ , or $n=1$ . From $n=3$ , it can fall to $n=2$ or $n=1$ . From $n=2$ , it can fall to $n=1$ . Each distinct jump emits a photon of a specific wavelength, producing a spectral line.

The total number of distinct spectral lines is the number of pairs of levels $(n_i, n_f)$ with $n_i > n_f$ , starting from $n=4$ down to $n=1$ . This is simply the number of combinations of 4 levels taken 2 at a time:

$\binom{4}{2} = \frac{4 \times 3}{2} = 6.$ …
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TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
$4 : 3$
›Reveal solutionSolution
The kinetic energy of an electron in a hydrogen atom is proportional to $1/n^2$ . The third excited state corresponds to $n=4$ and the fourth excited state to $n=5$ , so the ratio of their kinetic energies is $\frac{1/4^2}{1/5^2} = 25 : 16$ . The answer is (C).

Concept & Intuition

In the Bohr model of the hydrogen atom, an electron’s total energy is $E_n = -\frac{13.6}{n^2}$ eV. But kinetic energy is positive and exactly equal to the magnitude of the total energy (since $E = KE + PE$ , and $PE = -2 \times KE$ for a Coulomb force). So $KE_n = \frac{13.6}{n^2}$ eV.

The key: kinetic energy scales as $1/n^2$ . The “excited states” are counted starting from the ground state ( $n=1$ ) as the first. So the third excited state means $n=4$ , and the fourth excited state means $n=5$ . The ratio is then simply $(1/4^2) : (1/5^2) = 25 : 16$ .

Step-by-step reasoning

Identify the quantum numbers

Ground state: $n=1$ (first energy level)

First excited state: $n=2$

Second excited state: $n=3$

Third excited state: $n=4$

Fourth excited state: $n=5$

Watch out

A common mistake is to think “third excited” means $n=3$ . Remember: “excited” counts above the ground state, so the $k$ -th excited state has $n = k+1$ .

Write the kinetic energy formula

For hydrogen, the kinetic energy in the $n$ -th orbit is
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TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
$n^{\text{th}}$
›Reveal solutionSolution
The problem uses the Bohr radius formula $r_n = n^2 a_0$ and sets $r_{n+1} - r_n = r_{n-1}$ . Solving the resulting quadratic gives $n = 4$ , so the correct option is (C).

The key idea is that in the Bohr model, the radius of the $n$ -th orbit is proportional to $n^2$ . So the difference between successive radii grows as we go to higher orbits. The condition given — that this difference equals the radius of a lower orbit — sets up a simple quadratic in $n$ .

Let’s work through it step by step.

Write the radii in terms of $n$ . In the Bohr model, the radius of the $n$ -th orbit is

$r_n = n^2 a_0,$
where $a_0$ is the Bohr radius (a constant). So we have:
$r_{n} = n^2 a_0, \quad r_{n+1} = (n+1)^2 a_0, \quad r_{n-1} = (n-1)^2 a_0.$

Translate the given condition into an equation. The problem says:

$r_{n+1} - r_n = r_{n-1}.$
Substituting the expressions:
$(n+1)^2 a_0 - n^2 a_0 = (n-1)^2 a_0.$
Since $a_0 \neq 0$ , we can divide through by $a_0$ :
$(n+1)^2 - n^2 = (n-1)^2.$

Simplify the left-hand side.

$(n+1)^2 - n^2 = (n^2 + 2n + 1) - n^2 = 2n + 1.$
So the equation becomes:
$2n + 1 = (n-1)^2.$

Expand and solve the quadratic.

$2n + 1 = n^2 - 2n + 1.$
Cancel the $+1$ on both sides:
$2n = n^2 - 2n.$
Bring all terms to one side:
$0 = n^2 - 4n.$
Factor:
$n(n - 4) = 0.$
So $n = 0$ or $n = 4$ . …
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TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
$9 : 4$
›Reveal solutionSolution
Hydrogen-atom energy depends only on the principal quantum number: $E_n\propto -1/n^2$ . The first excited state is $n=2$ and the second excited state is $n=3$ , so the energy ratio is $|E_2|:|E_3| = \frac14:\frac19 = 9:4$ , option (A).

In the Bohr model the electron's energy is quantised:
$E_n=-\frac{13.6\ \text{eV}}{n^2},\qquad n=1,2,3,\dots$
The ground state is $n=1$ ; an "excited state" is any state above it. So the first excited state is $n=2$ and the second excited state is $n=3$ (a common error is to call $n=1$ the first excited state — that is the ground state).

Energies (magnitudes).

$|E_2|=\frac{13.6}{2^2}=\frac{13.6}{4},\qquad |E_3|=\frac{13.6}{3^2}=\frac{13.6}{9}.$

Ratio. $\frac{|E_2|}{|E_3|}=\frac{13.6/4}{13.6/9}=\frac{9}{4}.$ …
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TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQ
$Q. A hydrogen atom is excited from ground state to second excited state when it absorbs a photon. The energy of the photon is (A) 3.4 eV (B) 10.2 eV (C) 12.09 eV (D) 12.75 eV$
›Reveal solutionSolution
The key idea is that the energy of the absorbed photon equals the difference between the energy levels of the final and initial states. For hydrogen, the ground state is $n=1$ and the second excited state is $n=3$ . Using the formula $E_n = -13.6\,\text{eV}/n^2$ , the photon energy is $E_3 - E_1 = 12.09\,\text{eV}$ , so the correct option is (C).

The concept here is the Bohr model of the hydrogen atom, where electrons occupy discrete energy levels given by $E_n = -\frac{13.6\,\text{eV}}{n^2}$ , with $n=1,2,3,\dots$ . When an electron jumps from a lower to a higher energy level, it must absorb a photon whose energy exactly matches the difference between those levels. The trick is correctly identifying which $n$ corresponds to the "second excited state"—a common pitfall is confusing "excited state" numbering with the principal quantum number.

Let’s work through it step by step:

Identify the initial and final states.

The ground state is $n=1$ . The "second excited state" means the state with the second-highest energy above the ground state. The first excited state is $n=2$ , the second excited state is $n=3$ . So the electron goes from $n=1$ to $n=3$ .

Write the energy formula for hydrogen.

$E_n = -\frac{13.6\,\text{eV}}{n^2}$
This gives negative energies because the electron is bound; the more negative, the lower the energy.

Calculate the energy of the ground state.

$E_1 = -\frac{13.6}{1^2} = -13.6\,\text{eV}$

Calculate the energy of the second excited state.

$E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.511\,\text{eV}$

Find the energy difference (photon energy). The photon must supply exactly the gap: $\Delta E = E_3 - E_1 = (-1.511) - (-13.6) = 12.089\,\text{eV}$ …
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TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
$-0.85~\text{eV}$
›Reveal solutionSolution
The fourth excited state of the hydrogen atom corresponds to the principal quantum number $n=5$ . Using the Bohr model, the energy of an electron in this state is calculated as $-0.544~\text{eV}$ .

The energy levels of an electron in a hydrogen atom are quantized, meaning the electron can only exist at specific discrete energy values. These energy levels are described by the principal quantum number, $n$ , where $n$ can be any positive integer ( $1, 2, 3, \dots$ ).

The lowest possible energy state for an electron is called the ground state, which corresponds to $n=1$ . Any state with $n > 1$ is called an excited state. The excited states are numbered sequentially:

The first excited state corresponds to $n=2$ .

The second excited state corresponds to $n=3$ .

The third excited state corresponds to $n=4$ .

And so on.

In general, the $k$ -th excited state corresponds to the principal quantum number $n = k+1$ .
$n$
Let's apply this understanding to the problem.

Identify the principal quantum number ( $n$ ):

The question asks for the energy of an electron in the fourth excited state. Following our definition, the $k$ -th excited state corresponds to $n = k+1$ .

For the fourth excited state, $k=4$ , so the principal quantum number is $n = 4+1 = 5$ .

Watch out

A common mistake is to assume that the "fourth excited state" means $n=4$ . Remember that $n=1$ is the ground state, so the first excited state is $n=2$ , the second is $n=3$ , and the third is $n=4$ . Therefore, the fourth excited state must be $n=5$ .

Calculate the energy ( $E_n$ ):

Now, substitute $n=5$ into the energy formula for the hydrogen atom:

$E_5 = -\frac{13.6}{5^2}~\text{eV}$ …
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TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
$n = 3$
›Reveal solutionSolution
The Hα transition (n=3 → n=2) in hydrogen emits a photon of energy 1.89 eV. For photoelectron emission, the metal’s work function must be less than or equal to this photon energy, so the maximum work function is 1.89 eV, corresponding to option (C).

The key concept here is the photoelectric effect: a photon can eject an electron from a metal only if its energy is at least as large as the metal’s work function (the minimum energy needed to remove an electron). The Hα photon’s energy is fixed by the hydrogen energy levels, so the maximum work function a metal can have and still emit photoelectrons is exactly that photon energy.

We find the photon energy from the hydrogen atom’s energy levels:

Recall the hydrogen energy formula The energy of a level with principal quantum number $n$ is

$E_n = -\frac{13.6\ \text{eV}}{n^2}$
This comes from the Bohr model; the negative sign means the electron is bound.

Calculate the energies for n=3 and n=2

$E_3 = -\frac{13.6}{9} = -1.511\ \text{eV}$ $E_2 = -\frac{13.6}{4} = -3.4\ \text{eV}$

Find the photon energy from the transition The photon energy equals the difference between the initial and final levels:

$\Delta E = E_3 - E_2 = (-1.511) - (-3.4) = 1.889\ \text{eV}$
This is the Hα photon energy.

Apply the photoelectric condition …
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TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
$\lambda$
›Reveal solutionSolution
The key idea is that the energy difference between levels determines the emitted wavelength via $E = hc/\lambda$ . For the first transition, $2E - E = E = hc/\lambda$ . For the second, $\frac{5}{3}E - E = \frac{2}{3}E = hc/\lambda'$ . Since $\frac{2}{3}E$ is two-thirds of the first energy difference, the new wavelength is $\frac{3}{2}\lambda$ . The correct option is (B).

The relevant concept here is the inverse proportionality between photon energy and wavelength: $E_{\text{photon}} = \frac{hc}{\lambda}$ . When an atom drops from a higher energy level to a lower one, the emitted photon’s energy equals the difference between the two levels. So if we know the energy difference for one transition, we can scale it to find the wavelength for another.

Let’s work through it step by step.

First transition: from $2E$ to $E$ . The energy difference is

$\Delta E_1 = 2E - E = E.$
This photon has wavelength $\lambda$ , so
$E = \frac{hc}{\lambda}.$

Second transition: from $\frac{5}{3}E$ to $E$ . The energy difference is

$\Delta E_2 = \frac{5}{3}E - E = \frac{2}{3}E.$
Let the new wavelength be $\lambda'$ . Then
$\frac{2}{3}E = \frac{hc}{\lambda'}.$

Relate the two: From step 1, $E = hc/\lambda$ . Substitute this into step 2:

$\frac{2}{3} \cdot \frac{hc}{\lambda} = \frac{hc}{\lambda'}.$
Cancel $hc$ (nonzero) from both sides:
$\frac{2}{3\lambda} = \frac{1}{\lambda'}.$
Invert both sides:
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TG EAPCET 2021Set ap-2021-08-10-FN1 markMCQ
$\lambda$
›Reveal solutionSolution

Photon energy is inversely proportional to wavelength; the smaller energy gap $\tfrac{2}{3}E$ gives $\lambda'=\tfrac{3}{2}\lambda$ — option (B).

First transition: $2E \to E$ releases a photon of energy
$\Delta E_1 = 2E - E = E,\qquad \lambda = \frac{hc}{E}.$
Second transition: $\tfrac{5}{3}E \to E$ releases
$\Delta E_2 = \tfrac{5}{3}E - E = \tfrac{2}{3}E.$
Since $\lambda \propto \dfrac{1}{\Delta E}$ : …
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TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
$\Delta E$
›Reveal solutionSolution

The key idea is that increasing the orbital angular momentum by $\frac{h}{2\pi}$ corresponds to a one-step quantum jump ( $\Delta l = 1$ ), so the electron goes from $n=1$ to $n=2$ . The energy absorbed is the difference between the $n=2$ and $n=1$ levels, which is $10.2\ \text{eV}$ .

The orbital angular momentum of an electron in a hydrogen atom is quantized. For a given principal quantum number $n$ , the orbital angular momentum is $\sqrt{l(l+1)} \frac{h}{2\pi}$ , where $l$ can be $0, 1, \dots, n-1$ . But the problem gives a simpler clue: the increase is exactly $\frac{h}{2\pi}$ . That is the fundamental unit of angular momentum in quantum mechanics — the reduced Planck constant $\hbar = \frac{h}{2\pi}$ .

When the angular momentum increases by exactly one unit of $\hbar$ , the azimuthal quantum number $l$ must increase by 1. In the ground state ( $n=1$ ), the only possible $l$ is $0$ . So after absorbing energy, the electron jumps to a state where $l=1$ . The smallest $n$ that allows $l=1$ is $n=2$ . Therefore, the transition is from $n=1$ to $n=2$ .

Watch out

A common mistake is to think the angular momentum of the ground state is $\frac{h}{2\pi}$ (i.e., $l=1$ ). But for $n=1$ , $l=0$ , so the ground-state angular momentum is actually zero. The increase of $\frac{h}{2\pi}$ takes it from $0$ to $\hbar$ , which corresponds to $l=1$ , not $l=0$ to $l=2$ . …

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Q.Consider aiming a beam of free electrons towards free protons. When they scatter, an electron and a proton cannot combine to produce a H-atom,

Concept understanding — Photon Energy

Photon Energy: The Energy Carried by Light

The Intuition: Colour Determines Energy

The Planck–Einstein Relation

What This Tells You

A Concrete Example

Why this formula?

Why Photon Energy is $E = hf$ — The Reasoning Behind the Formula

The problem that forced the formula

Einstein's radical idea (1905)

The equivalent forms

Method: Pattern-Match to the "Two-Body Capture Needs a Third Body" Family of Problems

Steps

Q.Consider aiming a beam of free electrons towards free protons. When they scatter, an electron and a proton cannot combine to produce a H-atom,

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