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Question

In a hydrogen like ion, the energy difference between the $2^{nd}$ excitation energy state and ground is 108.8 eV. The atomic number of the ion is:

The correct answer is
3

Hydrogen-like Ion Atomic Number Calculation

The energy of an electron in the n-th state of a hydrogen-like ion with atomic number Z is given by the formula:

$E_n = -13.6 \times \frac{Z^2}{n^2}$ eV

Determine Quantum Numbers

  • The ground state corresponds to the principal quantum number, $n = 1$.
  • The first excitation state corresponds to $n = 2$.
  • The second excitation state corresponds to $n = 3$.

Calculate Energy Difference

The energy difference is given between the 2nd excitation state ($n=3$) and the ground state ($n=1$). We calculate the energy required to transition from the ground state to the second excited state:

  • Energy of the ground state ($n=1$): $E_1 = -13.6 \times \frac{Z^2}{1^2}$ eV
  • Energy of the 2nd excitation state ($n=3$): $E_3 = -13.6 \times \frac{Z^2}{3^2} = -13.6 \times \frac{Z^2}{9}$ eV
  • The energy difference ($\Delta E$) is the difference between the higher energy state and the lower energy state:

$\Delta E = E_3 - E_1 = \left(-13.6 \times \frac{Z^2}{9}\right) - \left(-13.6 \times Z^2\right)$

$\Delta E = -13.6 \times Z^2 \times \left(\frac{1}{9} - 1\right) = -13.6 \times Z^2 \times \left(-\frac{8}{9}\right)$

$\Delta E = 13.6 \times Z^2 \times \frac{8}{9}$ eV

Solve for Atomic Number (Z)

We are given that the energy difference is 108.8 eV.

Set the calculated energy difference equal to the given value:

$13.6 \times Z^2 \times \frac{8}{9} = 108.8$

Rearrange the equation to solve for $Z^2$:

$Z^2 = \frac{108.8 \times 9}{13.6 \times 8}$

Calculate the value:

$13.6 \times 8 = 108.8$

$Z^2 = \frac{108.8 \times 9}{108.8}$

$Z^2 = 9$

Take the square root to find Z:

$Z = \sqrt{9} = 3$

Conclusion

The atomic number of the hydrogen-like ion is 3.

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