Solution:
The energy of 1 photon is:
"E=\\frac{E_m}{N_A}=\\frac{4.79\u00d710^5 J\/mol}{6.022\u00d710^{23} mol^{-1}}=7.95\u00d710^{-19} J"
The energy of a photon is related to the wavelength as:
"E=\\frac{h\u00d7c}{\\lambda}"
Therefore:
"\\lambda=\\frac{h\u00d7c}{E}=\\frac{6.626\u00d710^{-34} J\u22c5s \\,\\,\u00d7\\,\\,3\u00d710^8 m\/s}{7.95\u00d710^{-19} J}=2.5\u00d710^{-7}m=250 nm"
Answer: 250 nm.
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