Q.16 Physics ⭐ 360° Option Elimination Active

NEET 2021 Physics Question 16 Solution

Topic: Dual Nature of Radiation and Matter · Official NTA Booklet Code M1

Question

The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of 3.3×10⁻³ watt will be: (h = 6.6×10⁻³⁴ J s)

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Verified Against NTA Final Answer Key

Option (3) : 10^16

📚 Dual Nature of Radiation and Matter Easy

📋 Given Parameters & Problem Setup

Wavelength (λ) 600 nm = 600 × 10^-9 m

Source Power (P) 3.3 × 10^-3 W

Planck Constant (h) 6.6 × 10^-34 J s

Speed of Light (c) 3 × 10^8 m/s

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The 30-Second Exam Hall Shortcut: ⚡ 2-Step Photon Flux: E = hc/λ, n = P/E

⏱️ Save ~20 seconds

$E = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{600 \times 10^{-9}} = 3.3 \times 10^{-19}\text{ J}$ . Number of photons per second: $n = \frac{P}{E} = \frac{3.3 \times 10^{-3}}{3.3 \times 10^{-19}} = 10^{16}\text{ photons/s}$ ! Instant 15-second calculation matching Option (3)!

1 Compute energy of one photon:

E = \frac{hc}{\lambda} = \frac{19.8 \times 10^{-26}}{600 \times 10^{-9}} = 3.3 \times 10^{-19}\text{ J}

2 Power is energy per second:

P = n E \implies n = \frac{P}{E}

3 Substitute:

n = \frac{3.3 \times 10^{-3}\text{ W}}{3.3 \times 10^{-19}\text{ J}} = 10^{16}\text{ photons/s}

4 Match Option (3).

🛡️

Why You Can 100% Trust This Shortcut: Canonical Planck-Einstein photon hypothesis and power-flux quantum relations.

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Calculating the Energy of a Single Photon

According to Planck's quantum theory of radiation, light consists of discrete energy quanta called photons. The energy $E$ carried by a photon of wavelength $\lambda = 600\text{ nm} = 6 \times 10^{-7}\text{ m}$ is:

E = \frac{hc}{\lambda} = \frac{(6.6 \times 10^{-34}\text{ J}\cdot\text{s}) \times (3 \times 10^8\text{ m/s})}{600 \times 10^{-9}\text{ m}} = \frac{1.98 \times 10^{-25}}{6 \times 10^{-7}} = 3.3 \times 10^{-19}\text{ J}

Step 2

Expressing Power in Terms of Photon Emission Rate

If the monochromatic source emits $n$ photons per second on average, the total energy radiated per second equals the source power $P$ :

P = n E = n \left(\frac{hc}{\lambda}\right)

Step 3

Solving for the Number of Photons per Second (n)

Substitute $P = 3.3 \times 10^{-3}\text{ W}$ and $E = 3.3 \times 10^{-19}\text{ J}$ :

n = \frac{P}{E} = \frac{3.3 \times 10^{-3}\text{ W}}{3.3 \times 10^{-19}\text{ J}} = 10^{-3 - (-19)} = 10^{16}\text{ photons/s}

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360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 1

✗ Incorrect

Verdict & Scientific Reason:

Result of an arithmetic error in exponent subtraction (calculating $10^{-3} / 10^{-21}$ ).

⚠️

Student Fallacy (Why candidates pick this trap): Miscalculating

hc

order of magnitude.

2 2

✗ Incorrect

Verdict & Scientific Reason:

Off by a factor of 10 from mishandling $600\text{ nm}$ conversion.

⚠️

Student Fallacy (Why candidates pick this trap): Using

\lambda = 6 \times 10^{-8}\text{ m}

instead of

6 \times 10^{-7}\text{ m}

3 3

✓ 100% Correct

Verdict & Scientific Reason:

Exact number of photons emitted per second: $n = 10^{16}$ .

💡

Why this is airtight: None; precise quantum physics computation.

4 4

✗ Incorrect

Verdict & Scientific Reason:

Off by a factor of 10 from inverted wavelength conversion.

⚠️

Student Fallacy (Why candidates pick this trap): Exponent subtraction error.

⚠️

NTA Trap Alert: Nanometer to Meter Exponent Trap

Taking $600\text{ nm} = 6 \times 10^{-8}\text{ m}$ instead of $6 \times 10^{-7}\text{ m}$ shifts the final exponent to $10^{17}$ or $10^{15}$ .

🛡️ How to avoid the -1 mark:

600\text{ nm} = 600 \times 10^{-9}\text{ m} = 6 \times 10^{-7}\text{ m}

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Core NEET Syllabus Standard

NCERT Physics · Dual Nature of Radiation and Matter Standard Benchmark

Core Scientific Principle Tested:

"Energy of a photon is $E = hc/\lambda$ . The photon flux emitted by a source of power $P$ is $n = P / (hc/\lambda)$ ."

Syllabus Focus: Photon Theory of Light

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Physics Question 16 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ 2-Step Photon Flux: E = hc/λ, n = P/E

Step-by-Step Rigorous Workflow & Mathematical Proof

Calculating the Energy of a Single Photon

Expressing Power in Terms of Photon Emission Rate

Solving for the Number of Photons per Second (n)

360° Option Elimination Engine Key Feature

NTA Trap Alert: Nanometer to Meter Exponent Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — Dual Nature of Radiation and Matter

Photocurrent and Stopping Potential Variation with Linearly Decreasing Light Power

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