An atom radiates 1.5 x \(10^{-19}\)J of energy when an electron jumps from one energy level to another. What is the wavelength of the emitted radiation? (Planck’s constant =6.6 x \(10^{-34}\)Js. Speed of light in air=3 x \(10^{8}ms^{-1}\)
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Correct Answer: Option A
Explanation:
\(E=\lambda v=\frac{hc}{\lambda}\Rightarrow\lambda=\frac{hc}{E}={6.6 \times10^{-34}\times3\times10^8}{1.5\times10^{-19}}\\=1.32\times10^{-6}\)
\(E=\lambda v=\frac{hc}{\lambda}\Rightarrow\lambda=\frac{hc}{E}={6.6 \times10^{-34}\times3\times10^8}{1.5\times10^{-19}}\\=1.32\times10^{-6}\)