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NEET CHEMISTRYStructure of AtomEasy

Question

The value of Planck's constant is 6.63×1034 J s6.63 \times 10^{-34} \text{ J s}. The velocity of light is 3.0×108 m s13.0 \times 10^8 \text{ m s}^{-1}. The closest value to the wavelength in nanometers of a quantum of light with a frequency of 8×1015 s18 \times 10^{15} \text{ s}^{-1} is:

A

2 \times 10^{-25}

B

5 \times 10^{-18}

C

4 \times 10^1

D

3 \times 10^7

Step-by-Step Solution

The relationship between the speed of light (cc), frequency (ν\nu), and wavelength (λ\lambda) is given by the equation: c=νλc = \nu \lambda Rearranging to solve for wavelength (λ\lambda): λ=cν\lambda = \frac{c}{\nu} Given: c=3.0×108 m s1c = 3.0 \times 10^8 \text{ m s}^{-1} ν=8×1015 s1\nu = 8 \times 10^{15} \text{ s}^{-1}

Calculation: λ=3.0×1088×1015 m\lambda = \frac{3.0 \times 10^8}{8 \times 10^{15}} \text{ m} λ=0.375×107 m\lambda = 0.375 \times 10^{-7} \text{ m} λ=3.75×108 m\lambda = 3.75 \times 10^{-8} \text{ m}

To convert the wavelength into nanometers (1 nm=109 m1 \text{ nm} = 10^{-9} \text{ m}): λ=3.75×108 m×109 nm1 m\lambda = 3.75 \times 10^{-8} \text{ m} \times \frac{10^9 \text{ nm}}{1 \text{ m}} λ=3.75×101 nm\lambda = 3.75 \times 10^1 \text{ nm} λ=37.5 nm\lambda = 37.5 \text{ nm}

The closest value among the options is 4×1014 \times 10^1 (which is 40).

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 Atomplancksconstantvelocityclosestwavelength

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