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Which is the correct relationship? (a). ...

Which is the correct relationship?
(a). `E_(1)` of `H=1//2E_(2)` of `He^(+)=1//3E_(3)`of `Li^(2+)=1//4E_(4)`of `Be^(3+)`
(b). `E_(1)(H)=E_(2)(He^(+))=E_(3)(Li^(2+))=E_(4)(Be^(3+))`
(c). `E_(1)(H)=2E_(2)(He^(+))=3E_(3)(Li^(2+))=4E_(4)(Be^(3+))`
(d). No relation

A

`E_(1) "of" H=1//2 E_(2) "of" He^(+)=1//3 E_(3) "of" Li^(2+)1//4 E_(4) "of" Be^(3+)`

B

`E_(1)(H)=E_(2)(He^(+))=E_(3)(Li^(2+))=E_(4)(Be^(3+))`

C

`E_(1)(H)=2E_(2)(He^(+))=3E_(3)(Li^(2+))=4E_(4)(Be^(3+))`

D

No relation

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the energies of the electrons in the hydrogen-like ions given in the question. The energy of the nth orbital (E_n) in a hydrogen-like atom is given by the formula: \[ E_n = -\frac{13.6 \, Z^2}{n^2} \] where: - \( E_n \) is the energy of the nth orbital, - \( Z \) is the atomic number, - \( n \) is the principal quantum number. Let's evaluate the energies for each of the ions mentioned: 1. **Hydrogen (H)**: - For hydrogen, \( Z = 1 \) and \( n = 1 \): \[ E_1(H) = -\frac{13.6 \times 1^2}{1^2} = -13.6 \, \text{eV} \] 2. **Helium ion (He\(^+\))**: - For helium ion, \( Z = 2 \) and \( n = 1 \): \[ E_1(He^+) = -\frac{13.6 \times 2^2}{1^2} = -54.4 \, \text{eV} \] 3. **Lithium ion (Li\(^{2+}\))**: - For lithium ion, \( Z = 3 \) and \( n = 1 \): \[ E_1(Li^{2+}) = -\frac{13.6 \times 3^2}{1^2} = -122.4 \, \text{eV} \] 4. **Beryllium ion (Be\(^{3+}\))**: - For beryllium ion, \( Z = 4 \) and \( n = 1 \): \[ E_1(Be^{3+}) = -\frac{13.6 \times 4^2}{1^2} = -216.0 \, \text{eV} \] Now, we can summarize the energies: - \( E_1(H) = -13.6 \, \text{eV} \) - \( E_2(He^+) = -54.4 \, \text{eV} \) - \( E_3(Li^{2+}) = -122.4 \, \text{eV} \) - \( E_4(Be^{3+}) = -216.0 \, \text{eV} \) Next, let's analyze the relationships given in the options: (a) \( E_1(H) = \frac{1}{2} E_2(He^+) = \frac{1}{3} E_3(Li^{2+}) = \frac{1}{4} E_4(Be^{3+}) \) (b) \( E_1(H) = E_2(He^+) = E_3(Li^{2+}) = E_4(Be^{3+}) \) (c) \( E_1(H) = 2E_2(He^+) = 3E_3(Li^{2+}) = 4E_4(Be^{3+}) \) (d) No relation From our calculations, we can see that the energies are not equal, and they do not satisfy the relationships in options (a), (b), or (c). Therefore, the correct answer is: **(d) No relation.**

To solve the problem, we need to analyze the energies of the electrons in the hydrogen-like ions given in the question. The energy of the nth orbital (E_n) in a hydrogen-like atom is given by the formula: \[ E_n = -\frac{13.6 \, Z^2}{n^2} \] where: - \( E_n \) is the energy of the nth orbital, - \( Z \) is the atomic number, - \( n \) is the principal quantum number. ...
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RESONANCE ENGLISH-NUCLEAR CHEMISTRY-PART -II
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