Check that the ratio k e² / G me mp is dimensionless. Look up a Table of Physical Constants and determine the

Exercises · 1.3

k e^2 / G m_e m_p

Rajasthan RbseTextbookSubjective· 3mImportance★★★★★est

22% · 15/67 Questions

✓ Free question

Concept understanding — Inverse Square Law Comparison

The Intuition: Why Does Light Get Dimmer So Fast?

Imagine you're standing near a campfire. You feel its warmth on your face. Now take ten steps back. Does the warmth feel half as strong? No — it feels much weaker, maybe a quarter as strong. That's not an accident. It's a pattern that shows up everywhere in physics: gravity, light, sound, electric fields, even radiation.

The reason is simple: as you move away from a source, the same amount of energy (or force) has to spread out over a larger area. And that area grows with the square of the distance.

The Core Idea in One Picture

Think of a light bulb at the centre of a balloon. As you inflate the balloon, the light hitting the inner surface spreads thinner and thinner. If you double the radius of the balloon, the surface area becomes four times larger. So each patch of the balloon gets only one-fourth the light.

That's the inverse square law in a nutshell: double the distance → one-fourth the intensity.

The Precise Statement

I \propto \frac{1}{r^2} \quad \text{or} \quad I = \frac{k}{r^2}

where:

$I$ = intensity (brightness, force per unit area, etc.)
$r$ = distance from the source
$k$ = a constant that depends on the source's strength

If you compare two distances $r_1$ and $r_2$ , the ratio of intensities is:

\frac{I_2}{I_1} = \left( \frac{r_1}{r_2} \right)^2

This is the inverse square law comparison — you compare how strong a quantity is at two different distances by taking the inverse ratio of the squares of those distances.

Why "Inverse Square" and Not Just "Inverse"?

Because the geometry of space is three-dimensional. The surface of a sphere is $4\pi r^2$ . As $r$ grows, the sphere's surface grows as $r^2$ . Whatever is radiating outward (light, sound, gravity) must pass through that entire surface. So the amount per unit area drops as $1/r^2$ .

If we lived in a flat, two-dimensional world, the law would be $1/r$ (like ripples on a pond). In one dimension, it would be constant. The inverse square law is a direct consequence of living in three dimensions.

The Comparison: What It Really Means

When you compare two situations, you're not calculating absolute intensity — you're finding the ratio. For example:

A star is 3 times farther away than another identical star. How much dimmer does it appear?

\frac{I_{\text{far}}}{I_{\text{near}}} = \left( \frac{1}{3} \right)^2 = \frac{1}{9}

The farther star is 9 times dimmer. Not 3 times — 9 times. That's the punch of the square.

Watch out

A common mistake: thinking "twice the distance means half the intensity." It's actually one-fourth. The square makes the drop much steeper than linear intuition suggests.

Where This Law Applies (and Where It Doesn't)

Applies to:

Light from a point source (a star, a bulb)
Sound from a small source (a speaker in open air)
Gravitational force between two masses
Electric force between two charges
Radiation from a radioactive point source

Does NOT apply to:

Light from a laser beam (it stays collimated)
Sound inside a pipe (it's guided)
Gravity inside a planet (the mass distribution changes the law)
Very large distances in cosmology (space itself is expanding)

The One-Line Takeaway

Important

Inverse square law comparison: When distance multiplies by $n$ , intensity divides by $n^2$ . Always compare using the square of the distance ratio, not the distance itself.

That's the whole idea. The rest is just practice applying it to different problems — but the geometry never changes.

Searches such as "inverse square law examples physics" and "intensity distance relationship formula" recur across gravitation, electrostatics, and optics topics in the NCERT/CBSE Class 11-12 Physics curriculum, since this same $1/r^2$ pattern underlies all of them. Comparing intensities at two distances is a very common numerical-question format in JEE Main and NEET.

Why this formula?

Inverse Square Law Comparison — Why the Formula Holds

The Inverse Square Law appears in physics wherever a quantity spreads out uniformly from a point source in three-dimensional space. The core idea is that the intensity (or field strength) decreases as the square of the distance from the source.

1. The Intuition: Spreading Over a Sphere

Imagine a point source emitting energy, light, sound, or gravitational force equally in all directions.

At a distance $r$ , the energy is spread uniformly over the surface area of a sphere of radius $r$ .
The surface area of a sphere is:

A = 4\pi r^2

If the total power (or flux) emitted by the source is $P$ , then the intensity $I$ (power per unit area) at distance $r$ is:

I = \frac{P}{4\pi r^2}

Key insight: The same total power is spread over a larger and larger area as $r$ increases. Hence, intensity is inversely proportional to $r^2$ .

2. Derivation for Gravitational Force (Newton's Law)

Newton’s law of gravitation states:

F = \frac{G M m}{r^2}

Why $1/r^2$ ?

The gravitational field lines from a point mass $M$ radiate outward uniformly.
The number of field lines crossing a sphere of radius $r$ is constant (conservation of flux).
The density of field lines (force per unit mass) at distance $r$ is:

g = \frac{GM}{r^2}

This is because the total flux $\Phi = 4\pi G M$ is spread over $4\pi r^2$ , giving:

g = \frac{\Phi}{4\pi r^2} = \frac{GM}{r^2}

Thus, the force on a test mass $m$ is $F = m g = \frac{G M m}{r^2}$ .

3. Derivation for Coulomb's Law (Electrostatics)

Coulomb’s law for electric force between two point charges $q_1$ and $q_2$ :

F = \frac{k q_1 q_2}{r^2}

Why $1/r^2$ ?

Electric field lines from a point charge $q$ radiate radially outward (or inward for negative charge).
Gauss’s law states that the total electric flux through a closed surface is proportional to the enclosed charge:

\oint \mathbf{E} \cdot d\mathbf{A} = \frac{q}{\varepsilon_0}

For a sphere of radius $r$ centered on the charge, the field is radial and constant in magnitude:

E \cdot 4\pi r^2 = \frac{q}{\varepsilon_0}

Therefore:

E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}

The force on a test charge $q_2$ is $F = q_2 E = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}$ .

4. Derivation for Light/Radiation Intensity

For a point source of light emitting power $P$ :

At distance $r$ , the power is spread over a sphere of area $4\pi r^2$ .
Illuminance (intensity) is:

I = \frac{P}{4\pi r^2}

Why not $1/r$ ?

In 2D (e.g., a line source), intensity falls as $1/r$ because the circumference of a circle is $2\pi r$ .
In 3D, the surface area grows as $r^2$ , so intensity falls as $1/r^2$ .

5. The Common Mathematical Reason

All inverse square laws arise from conservation of flux in three-dimensional space with isotropic emission. The geometry forces:

\text{Intensity} \propto \frac{1}{r^2}

because the area over which the quantity spreads is proportional to $r^2$ .

6. Key Exam Points

Quantity	Formula	Reason
Gravitational force	$F = \frac{G M m}{r^2}$	Flux of field lines over sphere area
Electric force	$F = \frac{k q_1 q_2}{r^2}$	Gauss’s law + spherical symmetry
Light intensity	$I = \frac{P}{4\pi r^2}$	Power spread over sphere surface
Sound intensity	$I = \frac{P}{4\pi r^2}$	Same geometric spreading

Important: The inverse square law holds only for point sources in 3D space with no absorption or reflection. For extended sources or non-isotropic emission, the law is modified.

7. Common Misconception

Not because "the force gets weaker with distance" — that’s vague.
The exact $1/r^2$ comes from the geometry of a sphere, not from any arbitrary assumption.
If space had 2 dimensions, the law would be $1/r$ ; if 4 dimensions, $1/r^3$ .

Final takeaway: The inverse square law is a geometric necessity for any conserved quantity spreading uniformly from a point in three-dimensional space. Memorize the formula, but understand the sphere-area argument — it’s the core reasoning for every exam question.

The key idea is the Inverse Square Law Comparison, which examines the relative strengths of two fundamental forces governed by inverse square laws: the electrostatic force and the gravitational force.

Dimensionless Check: We analyze the units of each term in the ratio $\frac{k e^2}{G m_e m_p}$ .
- Units of Coulomb's constant $k$ : $[\text{N m}^2 \text{C}^{-2}]$
- Units of elementary charge $e$ : $[\text{C}]$
- Units of gravitational constant $G$ : $[\text{N m}^2 \text{kg}^{-2}]$
- Units of mass $m_e, m_p$ : $[\text{kg}]$ The units of the numerator $k e^2$ are $[\text{N m}^2 \text{C}^{-2}] [\text{C}^2] = [\text{N m}^2]$ . The units of the denominator $G m_e m_p$ are $[\text{N m}^2 \text{kg}^{-2}] [\text{kg}] [\text{kg}] = [\text{N m}^2]$ . Since the units cancel out, the ratio is dimensionless.
Value Determination: Using standard physical constants:
- $k = 8.9875 \times 10^9 \text{ N m}^2 \text{C}^{-2}$
- $e = 1.602 \times 10^{-19} \text{ C}$
- $G = 6.674 \times 10^{-11} \text{ N m}^2 \text{kg}^{-2}$
- $m_e = 9.109 \times 10^{-31} \text{ kg}$ (electron mass)
- $m_p = 1.672 \times 10^{-27} \text{ kg}$ (proton mass) Substituting these values: $\frac{k e^2}{G m_e m_p} = \frac{(8.9875 \times 10^9 \text{ N m}^2 \text{C}^{-2}) (1.602 \times 10^{-19} \text{ C})^2}{(6.674 \times 10^{-11} \text{ N m}^2 \text{kg}^{-2}) (9.109 \times 10^{-31} \text{ kg}) (1.672 \times 10^{-27} \text{ kg})}$ $\frac{k e^2}{G m_e m_p} \approx \frac{2.307 \times 10^{-28}}{1.016 \times 10^{-67}} \approx 2.27 \times 10^{39}$
Significance: This ratio compares the strength of the electrostatic force to the gravitational force between an electron and a proton. Its extremely large value indicates that the electrostatic force is vastly stronger than the gravitational force at the atomic scale. This dominance of the electromagnetic force is why gravity is negligible in atomic and molecular interactions.

✓Final answer

k e^2 / G m_e m_p

The ratio $k e^2 / G m_e m_p$ is a dimensionless quantity that compares the strength of the electrostatic force to the gravitational force between an electron and a proton, and its value is approximately $\boxed{2.27 \times 10^{39}}$ .

The problem asks us to examine a specific ratio of fundamental constants, first by checking its dimensions, then by calculating its numerical value, and finally by interpreting its physical significance. This ratio is a direct comparison of the strengths of two fundamental forces: the electrostatic force and the gravitational force. Both forces follow an inverse square law, meaning their strength diminishes with the square of the distance between interacting particles.

The terms in the numerator, $k e^2$ , are related to the electrostatic interaction. $k$ is Coulomb's constant, and $e$ is the elementary charge. The terms in the denominator, $G m_e m_p$ , are related to the gravitational interaction. $G$ is the gravitational constant, $m_e$ is the mass of an electron, and $m_p$ is the mass of a proton. By forming this ratio, we are essentially comparing the "coupling strengths" of these two forces for a specific pair of particles (an electron and a proton).

Let's proceed step-by-step.

Identify the fundamental forces and their associated constants.

The ratio involves constants from two of the four fundamental forces:
- Electrostatic Force: Governed by Coulomb's Law, $F_e = k \frac{q_1 q_2}{r^2}$ . Here, $q_1 = e$ (charge of electron) and $q_2 = e$ (charge of proton, in magnitude). The relevant constants are Coulomb's constant $k$ and the elementary charge $e$ .
- Gravitational Force: Governed by Newton's Law of Universal Gravitation, $F_g = G \frac{m_1 m_2}{r^2}$ . Here, $m_1 = m_e$ (mass of electron) and $m_2 = m_p$ (mass of proton). The relevant constants are the gravitational constant $G$ , the electron mass $m_e$ , and the proton mass $m_p$ .
Determine the dimensions of each term in the ratio.

To check if the ratio is dimensionless, we need to find the dimensions of each constant and variable involved. We use $[M]$ for mass, $[L]$ for length, $[T]$ for time, and $[Q]$ for electric charge.
- Dimensions of $k$ (Coulomb's constant):
  
  From Coulomb's Law, $F_e = k \frac{q_1 q_2}{r^2}$ , we can write $k = \frac{F_e r^2}{q_1 q_2}$ .
  
  The dimensions of force are $[F] = [M L T^{-2}]$ .
  
  The dimensions of distance squared are $[r^2] = [L^2]$ .
  
  The dimensions of charge squared are $[q_1 q_2] = [Q^2]$ .
  
  Therefore, $[k] = \frac{[M L T^{-2}] [L^2]}{[Q^2]} = [M L^3 T^{-2} Q^{-2}]$ .
- Dimensions of $e^2$ (elementary charge squared):
  
  The dimension of elementary charge $e$ is $[Q]$ .
  
  Therefore, $[e^2] = [Q^2]$ .
- Dimensions of $G$ (gravitational constant):
  
  From Newton's Law of Gravitation, $F_g = G \frac{m_1 m_2}{r^2}$ , we can write $G = \frac{F_g r^2}{m_1 m_2}$ .
  
  The dimensions of force are $[F] = [M L T^{-2}]$ .
  
  The dimensions of distance squared are $[r^2] = [L^2]$ .
  
  The dimensions of mass squared are $[m_1 m_2] = [M^2]$ .
  
  Therefore, $[G] = \frac{[M L T^{-2}] [L^2]}{[M^2]} = [M^{-1} L^3 T^{-2}]$ .
- Dimensions of $m_e m_p$ (product of electron and proton mass):
  
  The dimension of mass is $[M]$ .
  
  Therefore, $[m_e m_p] = [M^2]$ .
Combine the dimensions to check if the ratio is dimensionless.

Now, let's substitute these dimensions into the given ratio:

\left[ \frac{k e^2}{G m_e m_p} \right] = \frac{[k] [e^2]}{[G] [m_e m_p]}

= \frac{[M L^3 T^{-2} Q^{-2}] [Q^2]}{[M^{-1} L^3 T^{-2}] [M^2]}

= \frac{[M L^3 T^{-2}]}{[M^{(-1+2)} L^3 T^{-2}]}

= \frac{[M L^3 T^{-2}]}{[M L^3 T^{-2}]}

= [M^{1-1} L^{3-3} T^{-2-(-2)}]

= [M^0 L^0 T^0]

Since all the dimensions cancel out, the ratio is indeed **dimensionless**.

4. Look up the numerical values of the physical constants.

We will use standard values for these fundamental constants:

* Coulomb's constant, $k = 8.9875 \times 10^9 \text{ N m}^2/\text{C}^2$

* Elementary charge, $e = 1.602 \times 10^{-19} \text{ C}$

* Gravitational constant, $G = 6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2$

* Mass of electron, $m_e = 9.109 \times 10^{-31} \text{ kg}$

* Mass of proton, $m_p = 1.673 \times 10^{-27} \text{ kg}$

Calculate the numerical value of the ratio. First, calculate the numerator $k e^2$ :

k e^2 = (8.9875 \times 10^9 \text{ N m}^2/\text{C}^2) \times (1.602 \times 10^{-19} \text{ C})^2

= (8.9875 \times 10^9) \times (2.566404 \times 10^{-38}) \text{ N m}^2

= 23.068 \times 10^{-29} \text{ N m}^2

= 2.3068 \times 10^{-28} \text{ N m}^2

Next, calculate the denominator $G m_e m_p$:

G m_e m_p = (6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2) \times (9.109 \times 10^{-31} \text{ kg}) \times (1.673 \times 10^{-27} \text{ kg})

= (6.674 \times 9.109 \times 1.673) \times 10^{(-11 - 31 - 27)} \text{ N m}^2

= (60.79 \times 1.673) \times 10^{-69} \text{ N m}^2

= 101.69 \times 10^{-69} \text{ N m}^2

= 1.0169 \times 10^{-67} \text{ N m}^2

Now, divide the numerator by the denominator:

\frac{k e^2}{G m_e m_p} = \frac{2.3068 \times 10^{-28} \text{ N m}^2}{1.0169 \times 10^{-67} \text{ N m}^2}

= \frac{2.3068}{1.0169} \times 10^{(-28 - (-67))}

= 2.2685 \times 10^{39}

Rounding to three significant figures, the value is $2.27 \times 10^{39}$.

6. Interpret the significance of the ratio.

The ratio $\frac{k e^2}{G m_e m_p}$ compares the strength of the electrostatic force to the gravitational force between an electron and a proton.

Consider the electrostatic force $F_e = k \frac{e \cdot e}{r^2}$ and the gravitational force $F_g = G \frac{m_e m_p}{r^2}$ between an electron and a proton separated by a distance $r$ .

The ratio of these forces is:

\frac{F_e}{F_g} = \frac{k \frac{e^2}{r^2}}{G \frac{m_e m_p}{r^2}} = \frac{k e^2}{G m_e m_p}

The calculated value of $2.27 \times 10^{39}$ means that the electrostatic force between an electron and a proton is approximately $2.27 \times 10^{39}$ times stronger than the gravitational force between them.

> [!IMPORTANT]
> This extremely large ratio highlights why gravity is negligible at the atomic and subatomic scales compared to the electromagnetic force. The structure of atoms and molecules is entirely determined by electromagnetic interactions, while gravity only becomes significant for objects with very large masses, like planets and stars.

> [!NOTE]
> This ratio is a fundamental constant of nature, independent of the distance between the particles, as the $r^2$ terms cancel out. It provides a profound insight into the relative strengths of two of the universe's fundamental interactions.

The ratio $k e^2 / G m_e m_p$ is dimensionless, and its value is approximately $2.27 \times 10^{39}$ . This ratio signifies that the electrostatic force between an electron and a proton is vastly stronger than the gravitational force between them, explaining why electromagnetic interactions dominate at the atomic scale.

✓Final answer

k e^2 / G m_e m_p

Method: Dimensional Analysis & Physical Constant Evaluation

Step 1 – Identify the dimensions of each constant

Constant	Symbol	SI Units	Dimensions
Coulomb constant	$k$	$\text{N·m}^2/\text{C}^2$	$[M L^3 T^{-4} A^{-2}]$
Elementary charge	$e$	C	$[A T]$
Gravitational constant	$G$	$\text{N·m}^2/\text{kg}^2$	$[M^{-1} L^3 T^{-2}]$
Electron mass	$m_e$	kg	$[M]$
Proton mass	$m_p$	kg	$[M]$

Step 2 – Write the dimensional formula of the ratio

The ratio is:

R = \frac{k e^2}{G m_e m_p}

Dimensions of numerator:

[k e^2] = [M L^3 T^{-4} A^{-2}] \cdot [A^2 T^2] = [M L^3 T^{-2}]

Dimensions of denominator:

[G m_e m_p] = [M^{-1} L^3 T^{-2}] \cdot [M] \cdot [M] = [M L^3 T^{-2}]

Step 3 – Check for dimensionless nature

\frac{[M L^3 T^{-2}]}{[M L^3 T^{-2}]} = [1]

The ratio is indeed dimensionless.

Step 4 – Evaluate the numerical value

From standard physical constants:

$k = 8.98755 \times 10^9 \ \text{N·m}^2/\text{C}^2$
$e = 1.60218 \times 10^{-19} \ \text{C}$
$G = 6.67430 \times 10^{-11} \ \text{N·m}^2/\text{kg}^2$
$m_e = 9.10938 \times 10^{-31} \ \text{kg}$
$m_p = 1.67262 \times 10^{-27} \ \text{kg}$

Numerator:

k e^2 = (8.98755 \times 10^9) \times (1.60218 \times 10^{-19})^2 \approx 2.307 \times 10^{-28}

Denominator:

G m_e m_p = (6.67430 \times 10^{-11}) \times (9.10938 \times 10^{-31}) \times (1.67262 \times 10^{-27}) \approx 1.017 \times 10^{-67}

Ratio:

\boxed{R \approx 2.27 \times 10^{39}}

Step 5 – Interpret the significance

This enormous number ( $\sim 10^{39}$ ) tells us:

Electrostatic force between a proton and an electron is $10^{39}$ times stronger than their gravitational attraction.
Gravity is extremely weak compared to electromagnetism at the particle scale.
This explains why atomic and molecular structures are governed by electric forces, not gravity — gravity only dominates for massive astronomical bodies.

Key insight: The ratio shows the relative strength of the two fundamental inverse-square forces (Coulomb vs. Newton) acting between the same two particles.

Common Mistakes: Inverse Square Law Ratio ( $k e^2 / G m_e m_p$ )

Mistake 1: Forgetting to check dimensions before plugging numbers

The error: Students rush to calculate the numerical value without first verifying the ratio is dimensionless. This leads to confusion about units and whether the answer "makes sense."

How to avoid: Always do a dimensional analysis first. Write each constant with its SI units:

$k$ (Coulomb constant): $\text{N·m}^2/\text{C}^2$
$e$ (electron charge): $\text{C}$
$G$ (gravitational constant): $\text{N·m}^2/\text{kg}^2$
$m_e$ , $m_p$ (masses): $\text{kg}$

Now check:

Numerator: $k e^2$ has units $(\text{N·m}^2/\text{C}^2) \times \text{C}^2 = \text{N·m}^2$
Denominator: $G m_e m_p$ has units $(\text{N·m}^2/\text{kg}^2) \times \text{kg}^2 = \text{N·m}^2$

Result: Both have same units → ratio is dimensionless. This confirms the numerical value is a pure number.

Mistake 2: Using wrong values from the constant table

The error: Students misread powers of 10 (e.g., writing $10^{-11}$ instead of $10^{-19}$ for $e$ ) or use inconsistent units (CGS vs SI).

How to avoid: Use SI units only and write each constant with its exponent clearly:

Constant	Symbol	Value (SI)
Coulomb constant	$k$	$8.99 \times 10^9$ N·m²/C²
Electron charge	$e$	$1.60 \times 10^{-19}$ C
Gravitational constant	$G$	$6.67 \times 10^{-11}$ N·m²/kg²
Electron mass	$m_e$	$9.11 \times 10^{-31}$ kg
Proton mass	$m_p$	$1.67 \times 10^{-27}$ kg

Pro tip: Write the ratio as:

\frac{k e^2}{G m_e m_p} = \frac{(8.99 \times 10^9)(1.60 \times 10^{-19})^2}{(6.67 \times 10^{-11})(9.11 \times 10^{-31})(1.67 \times 10^{-27})}

Mistake 3: Arithmetic errors with exponents

The error: Adding/subtracting exponents incorrectly when multiplying/dividing powers of 10.

How to avoid: Handle exponents separately from the coefficients:

Numerator exponent: $9 + 2(-19) = 9 - 38 = -29$
Denominator exponent: $-11 + (-31) + (-27) = -69$
Ratio exponent: $-29 - (-69) = 40$

So the ratio is roughly:

\frac{(8.99)(2.56)}{(6.67)(9.11)(1.67)} \times 10^{40}

Mistake 4: Not understanding what the ratio signifies

The error: Students calculate the number ( $\approx 2.3 \times 10^{39}$ ) but cannot explain its physical meaning.

How to avoid: Understand that this ratio compares two fundamental forces between an electron and a proton:

Electrostatic force: $F_e = k e^2 / r^2$
Gravitational force: $F_g = G m_e m_p / r^2$

The ratio $F_e / F_g = k e^2 / G m_e m_p$ (the $r^2$ cancels). So:

This ratio tells us the electrostatic force is about $10^{39}$ times stronger than the gravitational force between an electron and a proton.

Key insight: Gravity is extraordinarily weak compared to electromagnetism at the particle level. This is why atomic and molecular interactions are dominated by electric forces, not gravity.

Mistake 5: Thinking the ratio is universal for all particles

The error: Students assume this ratio applies to any two charged particles.

How to avoid: Remember this ratio is specific to electron-proton pairs. For other particle pairs (e.g., two protons), the masses change, so the ratio changes:

For two protons: $k e^2 / G m_p^2 \approx 1.2 \times 10^{36}$ (smaller, because protons are heavier)

Quick Summary Checklist

Step	What to do
✓	Check dimensions first
✓	Use SI units consistently
✓	Separate coefficients from exponents
✓	Interpret the ratio physically
✓	Note it's specific to electron-proton

←Previous question1.2 — The electrostatic force on a small sphere of charge 0.4muC due to anot…→Next question1.4 — (a) Explain the meaning of the statement 'electric charge of a body is…

Q.Check that the ratio ke2/Gmempk e^2 / G m_e m_pke2/Gme​mp​ is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

🎓Unlock everything free for 14 days

Q.Check that the ratio $k e^2 / G m_e m_p$ is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Unlock everything free for 14 days