Calculate the wavelength of the radiation emitted, when an excited atom of energy band gap 12.43 keV comes back to ground state.
Solution.
"\\Delta E = \\frac{h \\times c}{\\lambda}"
"\\lambda = \\frac{h \\times c}{\\Delta E}"
"\\lambda = \\frac{6.63*10^{-34} \\times 3*10^8}{1.99*10^{-15}} = 0.01 \\ nm"
Answer:
"\\lambda = 0.01 \\ nm"
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