Atomic StructureHard
Question
Photons of equal energy were incident on two different gas samples. One sample containing H-atoms in the ground state and the other sample containing H-atoms in some excited state with a principal quantum number ′n′. The photonic beams totally ionise the H-atoms. If the difference in the kinetic energy of the ejected electrons in the two different cases is 12.75 eV. Then find the principal quantum number ′n′ of the excited state.
Options
A.1
B.2
C.3
D.4
Solution
KE1 = Ephoton – BEn = 1
KE2 = Ephoton – BEn = n
KE2 - KE1 = BEn = 1 - BEn = n = 13.6 Z2
= 12.75 (given).
∴ n2 = 16 or n = 4.
BE : Binding energy.
KE2 = Ephoton – BEn = n
KE2 - KE1 = BEn = 1 - BEn = n = 13.6 Z2
= 12.75 (given).∴ n2 = 16 or n = 4.
BE : Binding energy.
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