Physics Q38 – JEE-MAIN 2026

Home Physics Dual Nature of MatterQuestion #75

A monochromatic source of light operating at

15 kW

emits

2.5 \times 10^{22}

photons/s. The region of an electromagnetic spectrum to which the emitted electromagnetic radiation belongs to _______. (Take

h = 6.6 \times 10^{- 34} J s

and

c = 3 \times 10^{8} m/s

Clue 1 of 3: Energy and Power Relationship

The total power emitted by the source is the total energy emitted per second. This power is also equal to the number of photons emitted per second multiplied by the energy of a single photon.

🧠 Full Solution Path

Step 1: Relate Power to Photon Energy and Rate✦ Active

The total power ( $P$ ) emitted by the source is the product of the number of photons emitted per second ( $N / t$ ) and the energy of a single photon ( $E_{p h o t o n}$ ). The energy of a single photon can be expressed in terms of its wavelength ( $λ$ ), Planck's constant ( $h$ ), and the speed of light ( $c$ ). Thus, we have:

P = (\frac{N}{t}) E_{p h o t o n} = (\frac{N}{t}) \frac{h c}{λ}

Rearranging the formula to solve for wavelength ( $λ$ ):

λ = \frac{(\frac{N}{t}) h c}{P}

Step 2: Calculate the Wavelength○ Expand

Substitute the given values into the formula:

$P = 15 kW = 15 \times 10^{3} W$

$N / t = 2.5 \times 10^{22} photons/s$

$h = 6.6 \times 10^{- 34} J s$

$c = 3 \times 10^{8} m/s$

λ = \frac{(2.5 \times 10^{22}) \times (6.6 \times 10^{- 34} J s) \times (3 \times 10^{8} m/s)}{15 \times 10^{3} W}

λ = \frac{49.5 \times 10^{- 4}}{15 \times 10^{3}} = 3.3 \times 10^{- 7} m

Converting to nanometers ( $1 m = 10^{9} nm$ ):

λ = 3.3 \times 10^{- 7} m \times \frac{10^{9} nm}{1 m} = 330 nm

💡 Teacher's Secret Hint

Ensure all units are consistent (SI units) before calculation.

Step 3: Identify the Region of the Electromagnetic Spectrum○ Expand

The calculated wavelength is $330 nm$ . We compare this value to the typical ranges of the electromagnetic spectrum:

- Visible light: $400 nm - 700 nm$

- Ultraviolet (UV) light: $10 nm - 400 nm$

- Infrared (IR) light: $700 nm - 1 mm$

- Microwave: $1 mm - 1 m$

Since $330 nm$ falls within the range of $10 nm - 400 nm$ , the emitted radiation belongs to the Ultraviolet region.

💡 Teacher's Secret Hint

Recall the approximate wavelength ranges for different parts of the electromagnetic spectrum.

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Physics Question 38 – JEE-MAIN 2026

Clue 1 of 3: Energy and Power Relationship

🧠 Full Solution Path