Using Planck's constant h = 6.63 times 10^ -34 Js and speed of light c = 3.0 times 10^8 ms^ -1 , the wavelength (in metres) of a photon with frequency 8 times 10^ 15 s^ -1 is closest to:

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NEET Chemistry QUESTION #7347

Question 1

Using Planck's constant \(h = 6.63 \times 10^{-34}\) Js and speed of light \(c = 3.0 \times 10^8\) ms\(^{-1}\), the wavelength (in metres) of a photon with frequency \(8 \times 10^{15}\) s\(^{-1}\) is closest to:

\(3 \times 10^7\) m
\(2 \times 10^{-25}\) m
\(5 \times 10^{-18}\) m
\(3.75 \times 10^{-8}\) m✔️

Correct Answer Explanation

Formula: \(\lambda = \dfrac{c}{\nu}\)

\(\lambda = \dfrac{3.0 \times 10^8}{8 \times 10^{15}} = 3.75 \times 10^{-8}\) m

This wavelength falls in the UV region of the electromagnetic spectrum. Always use \(\lambda = c/\nu\) for photon wavelength problems.

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