Final answer:
D) n=2 and n=5.The pair of quantum states of hydrogen with the biggest energy difference is n=2 and n=5, having an energy difference of 12.09 eV.
Step-by-step explanation:
The energy difference between quantum states in a hydrogen atom can be determined using the formula En = -13.6 eV/n², where n is the principal quantum number for the corresponding energy level. To find which of the given pairs of quantum states has the biggest energy difference, we need to apply this formula for each pair and compare the results. The energy difference ΔE between two states is ΔE = |Efinal - Einitial|.
Let's calculate and compare:
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- For n=1 and n=2: ΔE = |(-13.6 eV/1²) - (-13.6 eV/2²)| = 10.2 eV
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- For n=3 and n=4: ΔE = |(-13.6 eV/3²) - (-13.6 eV/4²)| = 1.51 eV
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- For n=4 and n=6: ΔE = |(-13.6 eV/4²) - (-13.6 eV/6²)| = 2.04 eV
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- For n=2 and n=5: ΔE = |(-13.6 eV/2²) - (-13.6 eV/5²)| = 12.09 eV
Comparing these energy differences, we can see that the pair n=2 and n=5 has the biggest energy difference, which is 12.09 eV.