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In hydrogen atom, if the difference in the energy of the electron in n=2 and n=3 orbits is E, the ionization energy of hydrogen atom is
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(b) Energy =0.0258 amu (K = constant)For removing an electron n1 = 1 to n2=∞∴Ionization energy = 7.2 E
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Question Text | In hydrogen atom, if the difference in the energy of the electron in n=2 and n=3 orbits is E, the ionization energy of hydrogen atom is |
Answer Type | Text solution:1 |
Upvotes | 150 |