Name the device used for measuring the mass of atoms and molecules.

NCERT Exemplar · Q21

Q. Name the device used for measuring the mass of atoms and molecules.

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Concept understanding — Least Count Precision

The Problem: How Fine Can You Measure?

Imagine you have a standard 15 cm ruler. You look at the markings — there's a line for every centimetre, and between those, smaller lines for every millimetre. Now try to measure the thickness of a single sheet of paper. You place it against the ruler. Does it line up exactly with a millimetre mark? Almost certainly not. It falls somewhere between two millimetre lines.

What do you report? You can't say "2.3 mm" because your ruler doesn't show tenths of a millimetre. The best you can honestly say is "about 2 mm" or "between 2 and 3 mm". That limitation — the smallest change in the quantity that your instrument can reliably detect and display — is its least count.

Note

Least count is not about how good you are at estimating. It is a property of the instrument itself. Even a perfect observer cannot extract more detail than the instrument's scale allows.

The Precise Definition

Least count is the smallest value of a physical quantity that can be measured accurately using a given instrument. It is the resolution of the measuring device.

For a simple scale (like a ruler, voltmeter, or thermometer) with equally spaced markings, the least count is:

\text{Least Count} = \frac{\text{Value of one main scale division}}{\text{Number of divisions on the vernier (or sub-scale)}}

For instruments without a vernier (like a plain ruler), the least count is simply the value of the smallest division on the scale.

Examples you already know:

Instrument	Smallest Division	Least Count
Metre ruler	1 mm	1 mm
Stopwatch (digital)	0.01 s	0.01 s
Laboratory thermometer	1 °C	1 °C
Screw gauge	0.01 mm	0.01 mm

Why This Matters in Exams

When you record a measurement, you must write it to the correct number of decimal places — the least count determines that. If a ruler has a least count of 1 mm, you cannot report a length as 12.35 cm. The correct recording is 12.3 cm or 12.4 cm (the last digit is uncertain, but it must be at the place of the least count).

Watch out

A common mistake: writing 2.50 cm when your ruler only shows millimetres. That implies you measured to 0.01 cm (0.1 mm), which you didn't. Write 2.5 cm — the last digit is at the tenths place, matching your least count of 0.1 cm.

The Core Intuition …

Why this formula?

Least Count & Precision: Why the Formula Holds

Let's build this from first principles — understanding why the least count formula works, not just memorizing it.

1. What is Least Count?

Least Count (LC) is the smallest measurement an instrument can reliably indicate.

Think of a ruler with 1 cm marks but no mm marks — you can't measure 0.5 cm precisely. The least count is 1 cm.

2. The Core Formula

For a scale-type instrument (ruler, vernier caliper, micrometer):

\text{Least Count} = \frac{\text{Value of 1 main scale division}}{\text{Number of divisions on the vernier/circular scale}}

Why this formula?

Reasoning step-by-step:

The main scale has fixed divisions (e.g., 1 mm each).
The vernier scale has $N$ divisions that exactly span $(N-1)$ main scale divisions.
So, 1 vernier division = $\frac{N-1}{N}$ main scale divisions.

The difference between 1 main scale division and 1 vernier division is:

\text{LC} = 1 \text{ MSD} - 1 \text{ VSD} = 1 - \frac{N-1}{N} = \frac{1}{N} \text{ MSD}

That's exactly the formula above.

3. Example: Vernier Caliper

Main scale: 1 mm per division
Vernier scale: 10 divisions covering 9 mm

Then:

\text{LC} = \frac{1 \text{ mm}}{10} = 0.1 \text{ mm}

Why 0.1 mm? Because the 10th vernier mark aligns with the 9th main scale mark — the smallest shift you can detect is 0.1 mm.

4. For Circular Scales (Micrometer Screw Gauge)

Same logic, different geometry:

\text{LC} = \frac{\text{Pitch}}{\text{Number of circular scale divisions}}

Pitch = distance moved by spindle in one full rotation.

Why? One full rotation moves the spindle by the pitch. If the circular scale has $N$ divisions, each division corresponds to $\frac{\text{pitch}}{N}$ linear movement.

5. Precision vs. Least Count

Precision is half the least count (or sometimes $\pm$ LC/2).

\text{Precision} = \pm \frac{\text{LC}}{2}

Why half? …

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The device used for measuring the mass of atoms and molecules is the mass spectrometer, which works by ionizing particles and separating them based on their mass-to-charge ratio.

Atoms and molecules are incredibly small, far too tiny to be weighed on any conventional balance. A typical analytical balance can measure masses down to about $0.1 \text{ mg}$ ( $10^{-7} \text{ kg}$ ), but the mass of a single carbon atom is approximately $2 \times 10^{-26} \text{ kg}$ . This vast difference in scale means we need a fundamentally different approach to determine their masses.

The core idea behind measuring atomic and molecular masses is not to "weigh" them directly, but to exploit their physical properties when subjected to external forces. Specifically, if we can give these particles an electric charge and then accelerate them, their path can be influenced by electric and magnetic fields. The extent to which their path is altered depends on their mass and charge. This principle allows us to effectively "sort" particles by their mass-to-charge ratio, and from this, deduce their individual masses.

Here's how the device works:

Ionization: The first step is to convert the neutral atoms or molecules into ions. This is typically done by bombarding them with high-energy electrons, which knocks off one or more electrons, creating positive ions. For example, an atom $X$ might become $X^+$ . This is crucial because only charged particles can be manipulated by electric and magnetic fields.
Acceleration: These newly formed ions are then accelerated through an electric field. This gives them a uniform kinetic energy, meaning they all move at high speeds. The kinetic energy ( $KE$ ) of an ion with charge $q$ accelerated through a potential difference $V$ is given by $KE = qV = \frac{1}{2}mv^2$ , where $m$ is the mass and $v$ is the velocity.
Deflection: The accelerated ions then enter a region with a strong magnetic field (and sometimes an electric field). This field exerts a force on the moving charged particles, causing them to deflect from their straight path. The amount of deflection depends on the ion's mass ( $m$ ), its charge ( $q$ ), and its velocity ( $v$ ). Lighter ions with higher charge-to-mass ratios ( $q/m$ ) will be deflected more significantly than heavier ions with lower charge-to-mass ratios.
$m/z$

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Concept: Mass Spectrometry

The device used for measuring the mass of atoms and molecules is the mass spectrometer.

Method: Time-of-Flight (TOF) Mass Spectrometry

This is one of the most common and intuitive methods. It works on a simple principle: ions with the same kinetic energy but different masses take different amounts of time to travel a fixed distance.

Steps:

Ionisation

The sample (atoms or molecules) is vaporised and then bombarded with high-energy electrons. This knocks off one or more electrons, turning the neutral atoms/molecules into positive ions (cations).

Example: $M + e^- \rightarrow M^+ + 2e^-$
Acceleration

These positive ions are passed through an electric field. The field gives all ions the same kinetic energy ( $KE$ ).

KE = \frac{1}{2}mv^2 = qV

where $q$ is the charge (usually $+1$), $V$ is the accelerating voltage, $m$ is mass, and $v$ is velocity.

3. Drift (Time-of-Flight)

The ions enter a long, evacuated tube (the "flight tube") of known length $L$ . Since all have the same $KE$ , lighter ions move faster and heavier ions move slower.

The time $t$ taken to reach the detector is:

$t = \frac{L}{v} = L \sqrt{\frac{m}{2qV}}$ …

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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-2026-05-09-AN1 markMCQ
$\mu$
›Reveal solutionSolution
The central bright spot is the Airy disk formed by Fraunhofer diffraction at the lens aperture; its radius in the focal plane is given by $1.22 \lambda f / D$ , yielding about $2.44\ \mu\text{m}$ , so the correct option is (D).

The key concept here is Fraunhofer diffraction through a circular aperture. When a plane wave passes through a circular opening (here, the lens itself acts as the aperture), the diffraction pattern on a screen at the focal plane consists of a central bright disk (the Airy disk) surrounded by concentric dark and bright rings. The radius of the first dark ring defines the size of the central bright spot.

Why this approach works:

The lens focuses the parallel beam to a point in its focal plane, but diffraction at the lens aperture spreads the light into a pattern. The angular radius of the first minimum is $\theta = 1.22 \lambda / D$ , where $D$ is the aperture diameter. Since the focal plane is at distance $f$ from the lens, the linear radius is $r = f \theta = 1.22 \lambda f / D$ . This is the standard result for the Airy disk radius.

Step-by-step solution:

Identify the given quantities

Wavelength: $\lambda = 500\ \text{nm} = 500 \times 10^{-9}\ \text{m}$

Focal length: $f = 20\ \text{cm} = 0.20\ \text{m}$

Aperture diameter: $D = 5\ \text{cm} = 0.05\ \text{m}$

Recall the formula for the radius of the central bright spot

For a circular aperture, the first minimum occurs at an angle $\theta$ satisfying

$\sin \theta \approx \theta = 1.22 \frac{\lambda}{D}$
(small-angle approximation is valid here because $\lambda \ll D$ ).

The linear radius on the focal plane is
$r = f \theta = 1.22 \frac{\lambda f}{D}$

Plug in the numbers

$r = 1.22 \times \frac{(500 \times 10^{-9})\ \text{m} \times 0.20\ \text{m}}{0.05\ \text{m}}$
Simplify step by step:
$\frac{500 \times 10^{-9} \times 0.20}{0.05} = \frac{500 \times 10^{-9} \times 0.20}{0.05}$
First, $0.20 / 0.05 = 4$ , so
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TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
$Q.If the focal lengths of the objective and eyepiece of a giant refracting telescope are 20 m and 1 cm respectively, then the angular magnification of the telescope is (A) 5 (B) 500 (C) 20 (D) 2000$
›Reveal solutionSolution
The angular magnification of a refracting telescope in normal adjustment is the ratio of the objective focal length to the eyepiece focal length. Here, $f_o = 20\ \text{m}$ and $f_e = 1\ \text{cm}$ , so $M = 2000$ .

The key idea is simple: a telescope makes distant objects appear larger by bending light through two lenses. The objective lens (big, long focal length) collects light and forms a real, inverted image at its focal plane. The eyepiece (short focal length) then acts like a magnifying glass to view that image. The angular magnification tells you how many times larger the angle subtended at the eye becomes compared to looking directly.

For a telescope in normal adjustment — where the final image is at infinity (relaxed eye viewing) — the angular magnification $M$ is given by the ratio of the focal lengths:
$M = \frac{f_o}{f_e}$
This formula is derived from geometry: the objective creates an image that subtends an angle $\theta_i \approx h / f_o$ (where $h$ is the image height), and the eyepiece, when the image is placed at its focal point, presents that same height at an angle $\theta_e \approx h / f_e$ to the eye. The ratio $\theta_e / \theta_i$ gives $f_o / f_e$ .

Now let’s apply it step by step.

Identify the focal lengths. The objective focal length is $f_o = 20\ \text{m}$ . The eyepiece focal length is $f_e = 1\ \text{cm}$ .

Convert to consistent units. Since the formula is a ratio, units must match. Convert everything to metres or centimetres — either works. Let’s use metres: $f_e = 1\ \text{cm} = 0.01\ \text{m}$ .

Plug into the magnification formula.

$M = \frac{f_o}{f_e} = \frac{20\ \text{m}}{0.01\ \text{m}} = 2000$
The units cancel, leaving a pure number. …
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TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
$Q. Which of the following statements is not correct regarding significant figures? (A) All the zeroes between two non-zero digits are significant, no matter where the decimal point is (B) The terminal zero(s) in a number without a decimal point are significant (C) The terminal zero(s) in a number with a decimal point are significant (D) To remove ambiguities in determining the number of significant figures, the best way is to report the measurement in the powers of 10$
›Reveal solutionSolution
The statement that terminal zeros in a number without a decimal point are always significant is incorrect. Such zeros are ambiguous and may or may not be significant. The incorrect statement is (B).

In scientific measurements, significant figures represent the precision of a measurement. They include all the digits that are known with certainty plus one estimated digit. Understanding the rules for significant figures is crucial for correctly reporting and interpreting experimental data, ensuring that the precision of a calculation reflects the precision of the input measurements.

Let's analyze each statement regarding significant figures:

Analyze statement (A): "All the zeroes between two non-zero digits are significant, no matter where the decimal point is"

This statement describes "sandwich zeros" or "captive zeros". These zeros are always considered significant because they are part of the measured value and contribute to its precision.

Example 1: In $2005 \text{ m}$ , the two zeros between $2$ and $5$ are significant. The number has $4$ significant figures.

Example 2: In $1.08 \text{ g}$ , the zero between $1$ and $8$ is significant. The number has $3$ significant figures. This statement is correct.

Analyze statement (B): "The terminal zero(s) in a number without a decimal point are significant"

This statement refers to trailing zeros in whole numbers. This is a common point of ambiguity in significant figures. When a number like $100 \text{ kg}$ is written, it's unclear whether the zeros are significant (meaning the measurement is precise to the nearest unit) or if they are just placeholders for the magnitude (meaning the measurement is only precise to the nearest hundred).

If $100 \text{ kg}$ was measured to the nearest kilogram, it would have $3$ significant figures.

If $100 \text{ kg}$ was measured to the nearest ten kilograms, it would have $2$ significant figures ( $1.0 \times 10^2 \text{ kg}$ ).

If $100 \text{ kg}$ was measured to the nearest hundred kilograms, it would have $1$ significant figure ( $1 \times 10^2 \text{ kg}$ ). Because these zeros may or may not be significant depending on the precision of the measurement, stating that they "are significant" (implying always significant) is incorrect. This ambiguity is precisely why scientific notation is preferred for such numbers. This statement is not correct.

Analyze statement (C): "The terminal zero(s) in a number with a decimal point are significant"

This statement refers to trailing zeros when a decimal point is present. When a decimal point is explicitly written, any trailing zeros are considered significant. This is because they indicate that the measurement was made to that level of precision. …
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TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
$0.079000$
›Reveal solutionSolution
The key idea is that trailing zeros after a decimal point are significant, so the measurement $0.079000$ m has 5 significant figures. The correct option is (C).

The concept here is significant figures — the digits in a number that carry meaningful information about its precision. The trick is that zeros can be tricky: leading zeros (like the ones before the 7) are never significant because they only set the decimal place, but trailing zeros after a decimal point are significant because they indicate the measurement was made to that level of precision. In $0.079000$ , the zeros after the 9 tell us the length was measured to the nearest millionth of a meter, not just to the nearest thousandth.

Let’s work through it step by step:

Identify all non-zero digits.

The digits 7 and 9 are non-zero, so they are always significant. That gives us 2 significant figures so far.

Check the zeros between non-zero digits.

There are no zeros between 7 and 9, so nothing to add here.

Examine the leading zeros.

The zeros before the 7 (the one immediately after the decimal and the one before it? Actually, $0.079000$ has one leading zero: the first zero after the decimal point. But wait — the number is $0.079000$ , so the digits are: 0 . 0 7 9 0 0 0. The first zero (the one before the decimal) is not a significant figure; it’s just a placeholder. The zero right after the decimal is also a leading zero — it only tells us the number is less than one-tenth. So these leading zeros are not significant.

Examine the trailing zeros after the decimal. …
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TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
$Q. The dimensions of four wires of the same material are given below. The increase in length is maximum in the wire of (A) Length 100 cm, Diameter 1 mm (B) Length 200 cm, Diameter 2 mm (C) Length 300 cm, Diameter 3 mm (D) Length 50 cm, Diameter 0.5 mm$
›Reveal solutionSolution
Since $\Delta L\propto L/d^{2}$ for the same material and load, compute $L/d^{2}$ for each wire: $100,\,50,\,33.3,\,200$ . The short, thin wire (50 cm, 0.5 mm) wins — option (D).

The concept first

Hooke's law in the language of materials is defined by Young's modulus:
$Y=\frac{\text{stress}}{\text{strain}}=\frac{F/A}{\Delta L/L}\qquad\Longrightarrow\qquad \Delta L=\frac{FL}{AY}.$
Read the formula physically:

Longer wire $\Rightarrow$ more elongation, because the same fractional stretch acts on a greater original length.

Thicker wire $\Rightarrow$ less elongation, because the load is shared over a bigger cross-section (lower stress).

$Y$ is a property of the material only, so with all four wires of the same material $Y$ cancels out of the comparison. The load $F$ is the same for all (that is implicit in the comparison).

For a circular wire, $A=\dfrac{\pi d^{2}}{4}$ , so
$\Delta L=\frac{FL}{\left(\frac{\pi d^2}{4}\right)Y}=\frac{4F}{\pi Y}\cdot\frac{L}{d^{2}} \qquad\Longrightarrow\qquad \boxed{\Delta L \propto \frac{L}{d^{2}}}$
Notice the square on the diameter — that is what makes thinness so powerful. Halving the diameter quadruples the stretch, whereas doubling the length only doubles it. Many students rank by length alone and pick (C); the $d^2$ is the trap.

Step-by-step

1. Set up the comparison quantity. Use consistent units within the ratio (cm and mm are fine, since we only compare the same combination across the four wires):
$k=\frac{L}{d^{2}} .$
2. Evaluate for each wire.

| Wire | $L$ (cm) | $d$ (mm) | $d^2$ | $k=L/d^2$ |

|---|---|---|---|---| …
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TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
$S = \dfrac{\sqrt{a b}}{c^3 d^4}$
›Reveal solutionSolution
The percentage error in a product/quotient of powers is the sum of each individual percentage error multiplied by the absolute value of its exponent. For $S = \frac{\sqrt{a b}}{c^3 d^4}$ , the result is 9%.

The key idea is that when a physical quantity is expressed as a product of powers of measured variables, the relative (or percentage) error in the result is the sum of the relative errors of each variable, each weighted by the magnitude of its exponent. This comes directly from the rules of error propagation using differentials.

For a relation $S = k \cdot a^p b^q c^r d^s$ (where $k$ is a constant), the fractional error is:
$\frac{\Delta S}{S} = |p| \frac{\Delta a}{a} + |q| \frac{\Delta b}{b} + |r| \frac{\Delta c}{c} + |s| \frac{\Delta d}{d}$
The absolute values are crucial — errors always add, never cancel, because each measurement's uncertainty contributes positively to the total uncertainty.

Let's apply this step by step.

Rewrite the given expression in power form.

$S = \dfrac{\sqrt{a b}}{c^3 d^4} = \dfrac{a^{1/2} b^{1/2}}{c^3 d^4} = a^{1/2} b^{1/2} c^{-3} d^{-4}$

The exponents are: $a$ has $+\frac12$ , $b$ has $+\frac12$ , $c$ has $-3$ , $d$ has $-4$ .

Identify the percentage errors given.

$\frac{\Delta a}{a} \times 100 = 2\%$ , $\frac{\Delta b}{b} \times 100 = 1\%$ , $\frac{\Delta c}{c} \times 100 = 1\%$ , $\frac{\Delta d}{d} \times 100 = 1\%$ .

Apply the error propagation formula.

The percentage error in $S$ is:

$\frac{\Delta S}{S} \times 100 = \left| \frac12 \right| \cdot 2\% + \left| \frac12 \right| \cdot 1\% + \left| -3 \right| \cdot 1\% + \left| -4 \right| \cdot 1\%$

Compute each term.

From $a$ : $\frac12 \times 2\% = 1\%$

From $b$ : $\frac12 \times 1\% = 0.5\%$

From $c$ : $3 \times 1\% = 3\%$

From $d$ : $4 \times 1\% = 4\%$

Sum them up.

$1\% + 0.5\% + 3\% + 4\% = 8.5\%$

Watch out

A common mistake is to forget the absolute value of the exponent. If you treat $c^{-3}$ as having exponent $-3$ and then add $-3 \times 1\% = -3\%$ , you'd get a wrong total. Errors always add in magnitude — the minus sign only tells you the direction of the effect, not that it reduces uncertainty.

Now, 8.5% is not among the given options. This means we need to check the standard convention used in such problems. In many exam contexts (especially in error analysis), the exponent on $d$ is taken as $4$ (not $-4$ ), and the percentage error contribution is $4 \times 1\% = 4\%$ . But wait — we already did that. The issue is that the problem likely expects the exponents to be treated as positive magnitudes only, and the sum $1 + 0.5 + 3 + 4 = 8.5$ still doesn't match any option.

Let's re-examine the expression: $S = \dfrac{\sqrt{a b}}{c^3 d^4}$ . The square root applies to the product $ab$ , so $\sqrt{ab} = (ab)^{1/2} = a^{1/2} b^{1/2}$ . That's correct.

Perhaps the intended interpretation is that the percentage errors are given as 2%, 1%, 1%, 1% and the formula for error uses the absolute values of the powers as coefficients. Then:

$a$ : exponent $\frac12$ , contribution $\frac12 \times 2 = 1$

$b$ : exponent $\frac12$ , contribution $\frac12 \times 1 = 0.5$ …
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Q.Name the device used for measuring the mass of atoms and molecules.

Concept understanding — Least Count Precision

The Problem: How Fine Can You Measure?

The Precise Definition

Why This Matters in Exams

The Core Intuition …

Why this formula?

Least Count & Precision: Why the Formula Holds

1. What is Least Count?

2. The Core Formula

Why this formula?

3. Example: Vernier Caliper

4. For Circular Scales (Micrometer Screw Gauge)

5. Precision vs. Least Count

Why half? …

Concept: Mass Spectrometry

Method: Time-of-Flight (TOF) Mass Spectrometry

Steps:

Common Mistakes: "Device for Measuring Mass of Atoms and Molecules"

✗ Mistake 1: Confusing with "weighing" devices

✗ Mistake 2: Confusing with other analytical instruments

✗ Mistake 3: Writing an incomplete or vague answer

✗ Mistake 4: Confusing with atomic mass measurement units

The concept first

Step-by-step

Q.Name the device used for measuring the mass of atoms and molecules.

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