Question

The ionization potential of hydrogen-like species is 217.6 eV. The number of neutrons in that species is:

• A. 5
• B. 4
• C. 3
• D. 2

Hint:

For hydrogen-like species: \[ E = 13.6 Z^2 \, \text{eV} \] Always identify \(Z\) first before solving nuclear composition.

Answer 1

The Correct Option is A

Step 1: Use formula for ionization energy of hydrogen-like species.
For hydrogen-like atoms: \[ E = 13.6 Z^2 \, \text{eV} \] where \(Z\) is atomic number. Given: \[ E = 217.6 \, \text{eV} \]

Step 2: Determine atomic number.
Substitute into formula: \[ 217.6 = 13.6 Z^2 \] Divide both sides: \[ Z^2 = \frac{217.6}{13.6} = 16 \] So, \[ Z = 4 \]

Step 3: Identify the element.
Atomic number \(Z = 4\) corresponds to beryllium (Be). So the species is hydrogen-like beryllium ion \(Be^{3+}\).

Step 4: Determine mass number relation.
To find neutrons: \[ \text{Neutrons} = A - Z \] We need mass number \(A\). Hydrogen-like ion implies common stable isotope of beryllium is: \[ A = 9 \]

Step 5: Calculate number of neutrons.
\[ \text{Neutrons} = 9 - 4 = 5 \]

Step 6: Final conclusion.
Thus, the number of neutrons in the species is: \[ \boxed{5} \]