(a) Find the radius of the n = 7 Bohr orbit of a triply ionized beryllium atom (
ID: 1299463 • Letter: #
Question
(a) Find the radius of the n = 7 Bohr orbit of a triply ionized beryllium atom (Be3+, Z = 4).
6.48025e-10 m is correct
(b) Is the energy required to raise an electron from the n = 7 state to the n = 8 state in Be3+ greater than, less than, or equal to the energy required to raise an electron in hydrogen from the n = 7 state to the n = 8 state?
(c) Verify your answer to part (b) by calculating the relevant energies.
Please Answer PART B and C and explain why if possible. Thank You!
EBe3+ = J EH = JExplanation / Answer
ACCORDING TO BOHR'S ATOMIC MODEL, THE RADIUS OF ANY ORBIT OF A UNI-ELECTRONIC SPECIE CAN BE CALCULATED EASILY BY RELATING IT WITH THAT OF A HYDROGEN ATOM. BY USING THE BOHR'S POSTULATE OF QUANTIZATION OF ANGULAR MOMENTUM, THE COULOMB'S LAW AND THE MECHANICS OF CIRCULAR OBITAL MOTION OF ELECTRONS, THE RADIUS OF FIRST ORBIT OF HYDROGEN ATOM CAN BE CALCULATED IN THE FORM OF VARIOUS CONSTANTS AND ITS VALUE IS 0.529 Angstrom, AND FOR ANY OTHER UNI-ELECTRONIC SPECIE,
r = 0.529 {(n^2)/Z}
So, here, radius,
r = 0.529 (49/4)
=> r = 6.48025 Angstrom
Now use conversion factor to convert it into metre.
1 Angstrom = {10^(-10)} m
So,
r = 6.48025 X {10^(-10)} m.
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