Express unified atomic mass unit in kg.

NCERT Exemplar · Q22

Q. Express unified atomic mass unit in kg.

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Concept understanding — Unit Conversion

Unit Conversion: Why 1 Metre and 100 Centimetres Are the Same Thing

Imagine you're measuring the length of your desk. You pull out a ruler marked in centimetres and find it's 120 cm long. Your friend, using a metre stick, says it's 1.2 m. You're both right — you've just used different units to describe the same physical length.

That's the core idea: unit conversion is the process of changing how you express a quantity without changing the quantity itself.

The Intuition: Same Quantity, Different Labels

Think of a pizza. Whether you call it "one pizza" or "8 slices," the amount of pizza hasn't changed. You've just used a different unit (pizza vs. slice) to describe it.

Similarly, 1 metre and 100 centimetres are the same length — just like 1 pizza and 8 slices are the same amount. The number changes (1 becomes 100, or 1 becomes 8), but the actual thing being measured stays identical.

Note

This is the most important idea to hold onto: conversion changes the number, not the quantity. If you ever feel like the quantity has changed, you've made a mistake.

The Precise Statement

Unit conversion is the multiplication of a quantity by a conversion factor — a fraction equal to 1 — that cancels the old unit and introduces the new one.

A conversion factor looks like this:

\frac{\text{new unit}}{\text{old unit}} = 1

For length: $\frac{1 \text{ m}}{100 \text{ cm}} = 1$ and $\frac{100 \text{ cm}}{1 \text{ m}} = 1$ .

Why are these fractions equal to 1? Because 1 metre is 100 centimetres. The numerator and denominator describe the same physical length, so their ratio is exactly 1.

How to Convert: The Only Rule You Need

Multiply by a conversion factor that cancels the unit you have and leaves the unit you want.

Let's convert 120 cm to metres:

Start with what you have: $120 \text{ cm}$
Choose the conversion factor that has "cm" in the denominator (to cancel it) and "m" in the numerator: $\frac{1 \text{ m}}{100 \text{ cm}}$
Multiply:

120 \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}} = \frac{120}{100} \text{ m} = 1.2 \text{ m}

The "cm" units cancel just like numbers do: $\frac{\text{cm}}{\text{cm}} = 1$ .

Tip

Always write the units explicitly. If the units don't cancel correctly, you've used the wrong conversion factor. This catches 90% of conversion mistakes.

The Reverse: Metres to Centimetres

Now convert 1.2 m to cm. This time, you want "cm" to remain and "m" to cancel. Use $\frac{100 \text{ cm}}{1 \text{ m}}$ :

1.2 \text{ m} \times \frac{100 \text{ cm}}{1 \text{ m}} = 1.2 \times 100 \text{ cm} = 120 \text{ cm}

Notice: when going from a larger unit (m) to a smaller unit (cm), the number gets larger (1.2 → 120). When going from smaller to larger, the number gets smaller (120 → 1.2). This is a useful sanity check.

Common Conversion Factors You'll Use

Quantity	Relationship	Conversion Factors
Length	1 m = 100 cm	$\frac{1 \text{ m}}{100 \text{ cm}}$ , $\frac{100 \text{ cm}}{1 \text{ m}}$
Mass	1 kg = 1000 g	$\frac{1 \text{ kg}}{1000 \text{ g}}$ , $\frac{1000 \text{ g}}{1 \text{ kg}}$
Time	1 h = 60 min	$\frac{1 \text{ h}}{60 \text{ min}}$ , $\frac{60 \text{ min}}{1 \text{ h}}$
Speed	1 km/h = $\frac{1000}{3600}$ m/s	$\frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}}$

Why this formula?

Dimensional Analysis: Why the Key Principles Hold

Dimensional Analysis is a powerful tool in physics and engineering that lets us check the consistency of equations, derive relationships, and convert units. But why does it work? Let's build the reasoning from the ground up.

1. The Core Idea: Physical Quantities Have Dimensions

Every physical quantity (like length, time, mass) can be expressed in terms of fundamental dimensions. The most common set in mechanics is:

L = Length
M = Mass
T = Time

For example:

Speed has dimensions $[L T^{-1}]$
Force has dimensions $[M L T^{-2}]$
Energy has dimensions $[M L^2 T^{-2}]$

Why this matters: Two quantities can only be meaningfully compared or equated if they have the same dimensions. You cannot add apples to oranges — and you cannot add length to time.

2. The Principle of Dimensional Homogeneity

The key formula that underpins everything is:

Every valid physical equation must be dimensionally homogeneous.

This means: the dimensions on the left-hand side must equal the dimensions on the right-hand side.

Why must this hold?

Consider an equation like:

v = u + at

Left side: $[v] = L T^{-1}$
Right side: $[u] = L T^{-1}$ , $[at] = (L T^{-2})(T) = L T^{-1}$

Both sides have dimensions $L T^{-1}$ . If they didn't match, the equation would be physically meaningless — you'd be comparing quantities that cannot be equal in any real experiment.

Reasoning: Physical laws describe relationships between measurable quantities. If the dimensions don't match, the equation cannot represent a real physical relationship, because the numerical value would depend on the arbitrary choice of units.

3. The Buckingham Pi Theorem: Why We Can Derive Relationships

This is the deeper mathematical reason. The Buckingham Pi Theorem states:

If a physical problem involves $n$ variables and $k$ fundamental dimensions, then it can be reduced to $n-k$ independent dimensionless groups (called $\pi$ groups).

Why does this work?

Imagine you have a relationship:

f(Q_1, Q_2, \dots, Q_n) = 0

where each $Q_i$ has dimensions. Because the equation must be dimensionally homogeneous, we can rearrange it into a function of dimensionless products only:

F(\pi_1, \pi_2, \dots, \pi_{n-k}) = 0

The reasoning: Dimensions act as constraints. Each fundamental dimension (M, L, T) gives one constraint. So if you have $n$ variables and $k$ constraints, you only have $n-k$ independent dimensionless combinations.

Example: For a simple pendulum, the period $T$ depends on length $L$ , mass $m$ , and gravity $g$ . That's 4 variables with 3 dimensions (M, L, T). So $4-3 = 1$ dimensionless group: $\pi = \frac{T^2 g}{L}$ . This tells us $T \propto \sqrt{L/g}$ without solving any differential equation.

4. Why We Can Convert Units Using Dimensional Analysis

The conversion factor formula:

\text{Value in new unit} = \text{Value in old unit} \times \left(\frac{\text{old unit}}{\text{new unit}}\right)^{\text{dimension exponent}}

Why this works: …

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The unified atomic mass unit is defined as exactly one-twelfth the mass of a carbon-12 atom; converting through Avogadro's number gives $u = 1.66054 \times 10^{-27}$ kg.

The unified atomic mass unit ( $u$ , sometimes written as amu or dalton) is the standard unit for expressing atomic and molecular masses. It provides a convenient scale because atomic masses in $u$ are close to their mass numbers, making calculations in chemistry and nuclear physics far more intuitive than working with kilograms directly.

The definition is precise: one unified atomic mass unit is exactly $\frac{1}{12}$ of the mass of a single neutral carbon-12 atom in its ground state. This choice anchors the atomic mass scale to a stable, abundant isotope.

Converting $u$ to kilograms

The conversion hinges on two experimentally determined constants: the molar mass of carbon-12 and Avogadro's number.

1. Start with the molar mass of carbon-12

By definition, the molar mass of $^{12}\text{C}$ is exactly $12 \text{ g/mol} = 0.012 \text{ kg/mol}$ . This means one mole of carbon-12 atoms has a mass of exactly $12$ grams.

2. Relate molar mass to individual atomic mass

One mole contains Avogadro's number of atoms:

N_A = 6.02214076 \times 10^{23} \text{ mol}^{-1}

Therefore, the mass of a single carbon-12 atom is:

m_{^{12}\text{C}} = \frac{0.012 \text{ kg}}{6.02214076 \times 10^{23}} = 1.99265 \times 10^{-26} \text{ kg}

3. Apply the definition of the unified atomic mass unit

Since $1 \, u = \frac{1}{12} m_{^{12}\text{C}}$ , we have:

$u = \frac{1.99265 \times 10^{-26}}{12} = 1.66054 \times 10^{-27} \text{ kg}$ …

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Let’s break this down step by step — first the concept, then the common mistakes, and finally how to avoid each.

Concept First: What is the unified atomic mass unit?

The unified atomic mass unit (symbol: u, also called amu or Da) is defined as:

1 u = $\frac{1}{12}$ of the mass of one atom of carbon-12.

The mass of one carbon-12 atom is exactly 12 u, by definition.

To express 1 u in kg, we use the known mass of a carbon-12 atom in kilograms:

Mass of one $^{12}\text{C}$ atom = $1.992647 \times 10^{-26} \ \text{kg}$ (approximately)

So:

1 \ \text{u} = \frac{1.992647 \times 10^{-26} \ \text{kg}}{12}

\boxed{1 \ \text{u} \approx 1.660539 \times 10^{-27} \ \text{kg}}

This is the standard value used in exams (often rounded to $1.66 \times 10^{-27} \ \text{kg}$ ).

Common Mistakes & How to Avoid Them

✗ Mistake 1: Using Avogadro’s number incorrectly

What students do:

They write:

1 \ \text{u} = \frac{1 \ \text{g}}{N_A}

and then convert g to kg incorrectly, or forget the conversion entirely.

Why it’s wrong:

While it’s true that $1 \ \text{u} = \frac{1 \ \text{g/mol}}{N_A}$ , this is a derived relation, not the definition. If you use it, you must:

Remember $N_A = 6.022 \times 10^{23} \ \text{mol}^{-1}$
Convert 1 g to $10^{-3}$ kg

How to avoid:

Always start from the definition (carbon-12 mass divided by 12). If you use Avogadro’s number, write the full conversion:

1 \ \text{u} = \frac{10^{-3} \ \text{kg/mol}}{6.022 \times 10^{23} \ \text{mol}^{-1}} \approx 1.66 \times 10^{-27} \ \text{kg}

✗ Mistake 2: Forgetting the factor of 12

What students do:

They take the mass of a carbon-12 atom and directly say that’s 1 u.

Why it’s wrong:

The mass of one $^{12}\text{C}$ atom is 12 u, not 1 u. So dividing by 12 is essential.

How to avoid:

Memorise the definition as a formula:

1 \ \text{u} = \frac{\text{mass of one } ^{12}\text{C atom}}{12}

✗ Mistake 3: Wrong power of 10

What students do:

They write $1.66 \times 10^{-26}$ kg or $1.66 \times 10^{-28}$ kg.

Why it’s wrong:

The correct exponent is -27. A common slip is to misplace the decimal when dividing $1.99 \times 10^{-26}$ by 12.

How to avoid: …

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

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›Reveal solutionSolution

$[h]=\text{J·s}$ , identical to the unit of angular momentum.

Planck's constant appears in $E=h\nu$ , so $[h]=\dfrac{[E]}{[\nu]}=\dfrac{\text{J}}{\text{s}^{-1}}=\text{J·s}=\text{kg·m}^2\text{s}^{-1}$ .

Angular momentum $L=mvr$ has units $\text{kg}\cdot(\text{m/s})\cdot\text{m}=\text{kg·m}^2\text{s}^{-1}=\text{J·s}$ . …
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[!TLDR]

Current sensitivity $=\theta/I$ ; since $\theta$ is dimensionless, its dimension is $[A^{-1}]$ .

Concept

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Solution …
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GUJCET 2024Set 131 markMCQ
$Q. Js is the unit of ________ physical quantity. (A) Angular momentum (B) Work function (C) Moment of Inertia (D) Rydberg constant$
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Angular momentum $L=I\omega$ has SI unit $\text{kg m}^2\text{s}^{-1} = \text{J·s}$ . …

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GUJCET 2023Set 091 markMCQ
$kg^{-1}s^{-2}A$
›Reveal solutionSolution
[!TLDR]

Reducing $\text{m}^2\text{V}^{-1}\text{s}^{-1}$ to fundamental units gives $\text{kg}^{-1}\text{s}^2\text{A}$ .

Concept

Mobility $\mu=\dfrac{v_d}{E}$ has units of $\dfrac{\text{m/s}}{\text{V/m}}=\text{m}^2\text{V}^{-1}\text{s}^{-1}$ . Convert the volt to base units to express $\mu$ in fundamental units.

Solution
$\text{V}=\frac{\text{J}}{\text{C}}=\frac{\text{kg}\,\text{m}^2\text{s}^{-2}}{\text{A}\,\text{s}}=\text{kg}\,\text{m}^2\text{A}^{-1}\text{s}^{-3}.$
Then …
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GUJCET 2023Set 091 markMCQ
$M^1L^1T^{-2}A^{-2}$
›Reveal solutionSolution
[!TLDR] $[L]=M^1L^2T^{-2}A^{-2}$ → (B).

Concept

Self-inductance is defined by $\varepsilon=-L\dfrac{dI}{dt}$ , so $L=\dfrac{\varepsilon\,dt}{dI}$ . (NCERT Electromagnetic Induction.)

Solution …
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GUJCET 2023Set 091 markMCQ
$Q. The earth takes 24 h to rotate once about its axis. How much time does the Sun takes to shift by 1 minute viewed from the earth. (A) 4 minutes (B) 40 s (C) 4 s (D) 40 minutes$
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[!TLDR]

The Sun moves 1° in 4 minutes, so 1 arcminute of apparent shift takes 4 seconds.

Concept

Earth's rotation makes the Sun appear to sweep $360^\circ$ in 24 hours; a small angular shift corresponds to a proportional time.

Solution

Angular rate of the Sun's apparent motion:

$\frac{360^\circ}{24\ \text{h}} = 15^\circ/\text{h} = \frac{1^\circ}{4\ \text{min}}$ …
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GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. The dimensional formula of electric flux is ___.$
(a) M^1 L^-3 T^-3 A^-1
(b) M^1 L^3 T^-3 A^-1
(c) M^-1 L^3 T^-3 A^-1
(d) M^1 L^3 T^3 A^-1
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Flux = E x area, i.e. (V/m) x m^2 = volt-metre; expressing volt in base units gives M L^3 T^-3 A^-1.

Electric flux Phi = E x A, with units (V/m)(m^2) = V.m.

…

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GUJCET 2021Set 151 markMCQ
$[M^{-1} L^{2} T^{-3} A^{1}]$
›Reveal solutionSolution

Electric potential is energy per unit charge.

Concept: …
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GUJCET 2020Set 071 markMCQ
$1^\circ$
›Reveal solutionSolution

$24$ hours per $360^\circ$ gives $4$ minutes per degree.

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GUJCET 2020Set 071 markMCQ
$V\cdot s\cdot A^{-1}$
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$1\,\text{H}=\text{Wb/A}=\text{V·s/A}$ ; $\text{Wb·s·A}^{-1}$ has an extra second and is not inductance. …

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GUJCET 2019Set 131 markMCQ
$M^0L^0T^0$
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Torsional constant has the dimensions of torque, $M^1L^2T^{-2}$ .

Concept: For a torsion spring, restoring torque $\tau = C\theta$ , so $C = \tau/\theta$ . Angle $\theta$ is dimensionless, hence $C$ has the dimensions of torque (energy).

Steps: …
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GUJCET 2019Set 131 markMCQ
$M^1L^2T^1Q^{-2}$
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The dimensional formula asked for is that of resistance (equivalently inductive reactance $\omega L$ ): $M^1L^2T^{-1}Q^{-2}$ .

Concept: Resistance $R=\dfrac{V}{I}$ . Expressing potential and current in terms of charge $Q$ : $V=\dfrac{\text{energy}}{Q}=\dfrac{ML^2T^{-2}}{Q}$ and $I=\dfrac{Q}{T}$ .

Steps:

$[R]=\dfrac{[V]}{[I]}=\dfrac{ML^2T^{-2}Q^{-1}}{Q\,T^{-1}}$ . …
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Q.Express unified atomic mass unit in kg.

Concept understanding — Unit Conversion

Unit Conversion: Why 1 Metre and 100 Centimetres Are the Same Thing

The Intuition: Same Quantity, Different Labels

The Precise Statement

How to Convert: The Only Rule You Need

The Reverse: Metres to Centimetres

Common Conversion Factors You'll Use

Why this formula?

Dimensional Analysis: Why the Key Principles Hold

1. The Core Idea: Physical Quantities Have Dimensions

2. The Principle of Dimensional Homogeneity

Why must this hold?

3. The Buckingham Pi Theorem: Why We Can Derive Relationships

Why does this work?

4. Why We Can Convert Units Using Dimensional Analysis

Why this works: …

Converting $u$ to kilograms

Concept: Unified Atomic Mass Unit (u) and its Relation to Carbon-12

Definition

Method: Direct Definition Method

Steps

Concept First: What is the unified atomic mass unit?

Common Mistakes & How to Avoid Them

✗ Mistake 1: Using Avogadro’s number incorrectly

✗ Mistake 2: Forgetting the factor of 12

✗ Mistake 3: Wrong power of 10

Concept

Solution …

Concept

Solution

Concept

Solution …

Concept

Solution

Q.Express unified atomic mass unit in kg.

Converting uuu to kilograms

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Converting $u$ to kilograms

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