Q.Which of the following electromagnetic radiations is used for viewing objects through fog?

Showing the 12 most recent of 42 on this concept.

1. CBSE 2026Set 55/1/11 markMCQ
   Q.Electromagnetic waves used in a diagnostic tool in medicine have a wavelength range (A) $1\ \text{nm to } 10^{-3}\ \text{nm}$ (B) $400\ \text{nm to } 1\ \text{nm}$ (C) $1\ \text{mm to } 700\ \text{nm}$ (D) $0.1\ \text{m to } 1\ \text{mm}$

   ›Reveal solutionSolution

   The question asks for the wavelength range of electromagnetic waves used in a medical diagnostic tool. X-rays, used in radiography and CT scans, have wavelengths from about $1\ \text{nm}$ down to $10^{-3}\ \text{nm}$. The correct option is (A).

   The key here is to connect the medical diagnostic tool to the specific type of electromagnetic wave it uses. The most common diagnostic tools in medicine that rely on electromagnetic waves are X-ray machines (for bone fractures, chest X-rays, CT scans) and MRI machines (which use radio waves). The question gives wavelength ranges, so we need to match the correct range to the tool.

   X-rays are the standard for imaging hard tissues like bones. Their wavelengths are extremely short — much shorter than visible light. Visible light ranges from about $700\ \text{nm}$ (red) to $400\ \text{nm}$ (violet). X-rays lie beyond ultraviolet, in the range of roughly $1\ \text{nm}$ down to $0.001\ \text{nm}$ ($10^{-3}\ \text{nm}$). This matches option (A) exactly.

   Let’s check the other options to be sure:

   1. Option (B): $400\ \text{nm to } 1\ \text{nm}$ — This covers the ultraviolet and part of the visible spectrum. Ultraviolet is used in sterilisation, not in routine diagnostic imaging. So this is incorrect.

   2. Option (C): $1\ \text{mm to } 700\ \text{nm}$ — This spans the infrared region. Infrared is used in thermal imaging (thermography), but that is not a primary diagnostic tool like X-rays. The question likely refers to the most common tool, so this is not the best answer.

   3. Option (D): $0.1\ \text{m to } 1\ \text{mm}$ — This is the microwave region. Microwaves are used in radar and communication, not in medical diagnostics (except for some niche applications like microwave ablation, but not as a standard diagnostic tool). So this is incorrect.

   Watch out

   A common mistake is to confuse the wavelength ranges of different EM waves. Remember: visible light is around $400$–$700\ \text{nm}$, X-rays are much shorter ($1\ \text{nm}$ and below), and radio waves are much longer (metres). Always anchor your memory to visible light as a reference point.

   Thus, the only range that fits the standard diagnostic tool (X-ray imaging) is option (A).

   ✓Final answer

   The correct option is (A) $1\ \text{nm to } 10^{-3}\ \text{nm}$.

2. CBSE 2026Set ANNUAL1 mark
   Q.Which electromagnetic wave has the highest energy?

   ›Reveal solutionSolution

   Energy of an EM wave is proportional to its frequency; gamma rays have the highest frequency.

   Photon energy $E=hf$, so the electromagnetic wave with the highest frequency has the highest energy. In the electromagnetic spectrum, gamma rays have the highest frequency (shortest wavelength).

   ✓Final answer

   Gamma rays have the highest energy.

3. CBSE 2026Set DS1 markMCQ
   Q.Electromagnetic waves of minimum frequency is:

   i) ultraviolet rays

   ii) $X$ rays

   iii) $\gamma$-rays

   iv) microwaves

   ›Reveal solutionSolution

   Microwaves have the longest wavelength (lowest frequency) of the four, so they are the minimum-frequency wave.

   Concept. In the electromagnetic spectrum, frequency increases (and wavelength decreases) in the order:

   $\text{radio} < \text{microwave} < \text{infrared} < \text{visible} < \text{ultraviolet} < \text{X-rays} < \gamma\text{-rays}.$

   Among the given options, microwaves sit far to the low-frequency end, while $\gamma$-rays are highest. Hence the electromagnetic wave of minimum frequency here is the microwave.

   ✓Final answer

   (iv) microwaves.

4. CBSE 2026Set ANNUAL1 markMCQ
   Q.Which of the following has the smallest wavelength?

   (a) Micro-waves

   (b) Red light

   (c) Ultraviolet radiation

   (d) gamma-rays

   ›Reveal solutionSolution

   Across the electromagnetic spectrum, wavelength decreases (and frequency/energy increases) in the order: micro-waves > red light (visible) > ultraviolet > gamma-rays.

   The electromagnetic spectrum, arranged from LONGEST to SHORTEST wavelength, runs roughly: radio waves > microwaves > infrared > visible light (red to violet) > ultraviolet > X-rays > gamma rays. Comparing the four given options in that order - micro-waves (longest here), red light (visible, shorter than microwaves), ultraviolet (shorter still), and gamma-rays (shortest of all, typically less than about 10^-12 m, produced by nuclear transitions) - gamma rays clearly have the smallest wavelength.

   ✓Final answer

   (d) gamma-rays.

5. CBSE 2026Set ANNUAL1 markMCQ
   Q.Identify the constituent radiation of electromagnetic spectrum which is used in medicine to destroy cancer cells.

   (a) X-rays

   (b) UV rays

   (c) IR rays

   (d) Gamma rays

   ›Reveal solutionSolution

   Gamma rays, the highest-energy, shortest-wavelength part of the EM spectrum, are used in radiotherapy to destroy cancer cells.

   Among the options, X-rays are chiefly used for imaging (and in some radiotherapy), UV rays for sterilisation and detecting fluorescence, and IR rays for heating and thermal imaging. Gamma rays, emitted by radioactive nuclei, carry very high photon energy and deep penetrating power. In medicine this is exploited in gamma-ray therapy (e.g. Cobalt-60 teletherapy, the 'gamma knife'), where focused gamma radiation is used to destroy malignant (cancer) cells while limiting damage to surrounding healthy tissue.

   ✓Final answer

   (d) Gamma rays are used in medicine to destroy cancer cells.

6. CBSE 2025Set 55/6/11 markMCQ
   Q.Assertion (A): X-rays are produced when slow moving electrons are stopped by a metal target of high atomic number. Reason (R): X-rays consist of low-energy photons. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Both Assertion (A) and Reason (R) are false.

   ›Reveal solutionSolution

   Both statements are false: X-rays require FAST (not slow) electrons to be decelerated by a high-Z target, and X-ray photons are HIGH-energy (not low-energy). The correct option is (D).

   Concept and intuition

   X-rays are electromagnetic radiation of very short wavelength (roughly $10^{-11}$ m to $10^{-8}$ m) and correspondingly very HIGH photon energy — from about $100\,\text{eV}$ up to $100\,\text{keV}$, thousands of times more energetic than visible-light photons ($\sim2$–$3\,\text{eV}$). They are produced in an X-ray tube by bremsstrahlung ("braking radiation"): electrons are first accelerated through a large potential difference (tens of kilovolts), giving them very HIGH kinetic energy, and are then suddenly decelerated on striking a metal target of high atomic number (e.g. tungsten, $Z=74$) — a high-$Z$ target is used because the efficiency of bremsstrahlung production increases with $Z$.

   Evaluating the Assertion

   The Assertion states that X-rays are produced when slow moving electrons are stopped by a high-$Z$ target. This gets the mechanism backwards: it is precisely because the electrons are moving very fast (having been accelerated through a large voltage) that stopping them suddenly releases enough energy to produce X-ray photons. A genuinely slow-moving electron carries far too little kinetic energy to produce a photon anywhere in the X-ray range. So the Assertion, as literally worded, is false — the single word "slow" (which should read "fast") is what makes it false; the rest of the statement (high-Z target, sudden stopping) is correct in isolation, but the assertion as a whole is not.

   Evaluating the Reason

   The Reason states that X-rays consist of low-energy photons. This is also the opposite of the truth: X-ray photons are HIGH-energy compared to visible, infrared, or radio-wave photons — that is exactly why they can penetrate soft tissue and are dangerous in large doses. So the Reason is false as well.

   Watch out

   A common trap in this kind of assertion–reason question is to focus only on the correct half of the Assertion (a high-$Z$ target stopping electrons) and overlook the single wrong qualifier — "slow" should be "fast." A question's assertion must be judged exactly as written; do not mentally "correct" a wrong word and then treat the assertion as true. Always check every qualifier, since a deliberately inserted wrong word like this is a favourite CBSE trick.

   Tip

   Remember the Duane–Hunt law: $\lambda_{\min} = \dfrac{hc}{eV}$, where $V$ is the tube's accelerating voltage. A larger $V$ (faster electrons) gives a smaller $\lambda_{\min}$ (higher photon energy). "Slow" electrons (small $V$) would only give long-wavelength, low-energy photons — not genuine X-rays.

   Since both the Assertion and the Reason are false, the correct option is (D).

   ✓Final answer

   Both Assertion (A) and Reason (R) are false. The correct option is (D).

7. CBSE 2025Set JS1 markMCQ
   Q.Electro-magnetic wave of minimum frequency is: (A) Ultraviolet rays (B) X-rays (C) Gamma ($\gamma$-) rays (D) Micro waves

   ›Reveal solutionSolution

   In the EM spectrum the order of increasing frequency is microwave < ultraviolet < X-ray < $\gamma$-ray, so microwaves have the minimum frequency — option (D).

   Concept. For all electromagnetic waves $c=\nu\lambda$, so a larger wavelength means a smaller frequency. Arranged by increasing frequency:

   $\text{radio} < \text{micro} < \text{infrared} < \text{visible} < \text{UV} < \text{X-ray} < \gamma\text{-ray}.$

   Comparison of the options. Microwaves ($\sim10^{9}\text{–}10^{11}$ Hz) sit far below ultraviolet ($\sim10^{15}$ Hz), X-rays ($\sim10^{18}$ Hz) and $\gamma$-rays ($>10^{19}$ Hz). Hence microwaves have the lowest frequency.

   ✓Final answer

   Option (D) Micro waves.

8. CBSE 2025Set A1 mark
   Q.Write True or False: Frequency is maximum for long radio waves.

   ›Reveal solutionSolution

   The statement is False: frequency is minimum, not maximum, for long radio waves.

   In the electromagnetic spectrum, wavelength and frequency are inversely related (c = νλ). Radio waves have the longest wavelengths and hence the LOWEST frequencies (as low as a few Hz to MHz), while at the other end, gamma rays have the shortest wavelengths and the HIGHEST frequencies (up to 10²⁰ Hz or more). So it is incorrect to say frequency is maximum for long radio waves — it is in fact minimum.

   ✓Final answer

   False.

9. CBSE 2025Set A1 mark
   Q.Write answer in one sentence: What is the degree of order of frequency of visible light?

   ›Reveal solutionSolution

   The frequency of visible light is of the order of 10¹⁴ Hz.

   Visible light has wavelengths roughly between 400 nm (violet) and 700 nm (red). Using $c = \nu\lambda$ with $c = 3\times10^8$ m/s:

   For red light ($\lambda \approx 700$ nm): $\nu = \dfrac{3\times10^8}{700\times10^{-9}} \approx 4.3\times10^{14}$ Hz

   For violet light ($\lambda \approx 400$ nm): $\nu = \dfrac{3\times10^8}{400\times10^{-9}} \approx 7.5\times10^{14}$ Hz

   So across the visible spectrum, frequency ranges roughly from 4×10¹⁴ Hz to 7.5×10¹⁴ Hz — i.e., the degree/order of frequency of visible light is 10¹⁴ Hz.

   ✓Final answer

   Of the order of 10¹⁴ Hz.

10. CBSE 2025Set ANNUAL1 markMCQ
    Q.The frequencies of gamma-rays, ultraviolet rays, and X-rays are ν1, ν2 and ν3 respectively. Then

    (i) ν1 = ν2 = ν3

    (ii) ν1 > ν3 > ν2

    (iii) ν1 > ν2 > ν3

    (iv) ν3 > ν2 > ν1

    ›Reveal solutionSolution

    In the electromagnetic spectrum, frequency increases in the order: ultraviolet < X-rays < gamma rays.

    The electromagnetic spectrum is ordered by wavelength/frequency as follows (increasing frequency): radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays. Gamma rays (emitted in nuclear transitions) have the shortest wavelength and hence the highest frequency; X-rays (produced by high-energy electron transitions/deceleration) come next; ultraviolet rays have a comparatively lower frequency than both.

    Given $\nu_1$ = gamma rays, $\nu_2$ = ultraviolet rays, $\nu_3$ = X-rays, the correct ordering is:

    $u_1 > u_3 > u_2$

    ✓Final answer

    (ii) $\nu_1 >\nu_3 >\nu_2$ (gamma > X-ray > ultraviolet).

11. CBSE 2025Set ANNUAL1 markMCQ
    Q.Which has the longest wavelength ?

    (a) X-rays

    (b) Micro waves

    (c) Infrared rays

    (d) Radio waves

    ›Reveal solutionSolution

    In the electromagnetic spectrum, wavelength increases in the order X-rays < infrared < microwaves < radio waves, so radio waves are longest.

    The electromagnetic spectrum, arranged by increasing wavelength (decreasing frequency/energy), runs roughly:

    Gamma rays < X-rays < Ultraviolet < Visible light < Infrared < Microwaves < Radio waves.

    Among the four options, X-rays have the shortest wavelength (highest energy, used for medical imaging), infrared is longer, microwaves longer still, and radio waves have the longest wavelength of all electromagnetic radiation (from a few mm to several km).

    ✓Final answer

    Radio waves have the longest wavelength — option (d).

12. CBSE 2025Set ANNUAL1 markMCQ
    Q.The frequency of gamma-rays, x-rays and ultra-violet rays are a, b and c respectively. Then –

    (a) a < b < c

    (b) a > b > c

    (c) a = b = c

    (d) a > c > b

    ›Reveal solutionSolution

    In the electromagnetic spectrum, gamma rays have the highest frequency, then X-rays, then ultraviolet.

    Ordering the electromagnetic spectrum by increasing frequency (equivalently, decreasing wavelength): radio waves < microwaves < infrared < visible < ultraviolet < X-rays < gamma rays.

    So gamma-ray frequency $a$ is the highest, X-ray frequency $b$ is next, and ultraviolet frequency $c$ is the lowest of the three: $a > b > c$.

    ✓Final answer

    $a > b > c$ (option b).