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NEET CHEMISTRYStructure of AtomEasy

Question

Calculate the energy in Joule corresponding to light of wavelength 45 nm: (Planck's constant h = 6.63 × 10⁻³⁴ Js; speed of light c = 3 × 10⁸ ms⁻¹)

A

6.67 × 10¹⁵

B

6.67 × 10¹¹

C

4.42 × 10⁻¹⁵

D

4.42 × 10⁻¹⁸

Step-by-Step Solution

The energy (EE) of a photon is related to its wavelength (λ\lambda) by the Planck's equation: E=hcλE = \frac{hc}{\lambda}.

  1. Identify Given Values: Planck's constant (hh) = 6.63×10−34 Js6.63 \times 10^{-34} \text{ Js} Speed of light (cc) = 3×108 ms−13 \times 10^8 \text{ ms}^{-1}
  • Wavelength (λ\lambda) = 45 nm=45×10−9 m45 \text{ nm} = 45 \times 10^{-9} \text{ m}
  1. Calculation: E=(6.63×10−34)×(3×108)45×10−9E = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^8)}{45 \times 10^{-9}} E=19.89×10−2645×10−9E = \frac{19.89 \times 10^{-26}}{45 \times 10^{-9}} E=0.442×10−17 JE = 0.442 \times 10^{-17} \text{ J} E=4.42×10−18 JE = 4.42 \times 10^{-18} \text{ J}

Thus, the energy corresponds to Option D.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Structure of Atom. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYStructure of Atomcalculateenergycorrespondingwavelengthplancks

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