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Step-by-Step Solution
Step 1: Identify the Energy of the Hydrogen Transition (n = 2 → n = 1)
For a hydrogen atom (Z = 1), the energy of a transition from an initial state n2 to a final state n1 is given by the formula:
$E = 13.6 \times Z^2 \left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)\,$eV
Here, Z = 1, n1 = 1, and n2 = 2. Substituting these values:
$E_\text{H} = 13.6 \times 1^2 \left(\dfrac{1}{1^2} - \dfrac{1}{2^2}\right)\,$eV
$E_\text{H} = 13.6 \times \left(1 - \dfrac{1}{4}\right)\,$eV
$E_\text{H} = 13.6 \times \dfrac{3}{4}\,$eV = 10.2 eV
Thus, the radiation from the hydrogen transition n = 2 to n = 1 carries an energy of 10.2 eV.
Step 2: Examine Possible Transitions in the He+ Ion
For a He+ ion, the atomic number Z = 2. The energy for a transition from n1 to n2 is:
$E = 13.6 \times Z^2 \left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)\,$eV
Here, Z = 2, so the factor 13.6 × 22 = 13.6 × 4 = 54.4 eV multiplies the bracketed term. We need to check which of the given transitions matches the 10.2 eV from the hydrogen emission. The main possibilities given are:
n = 1 → n = 4
n = 2 → n = 3
n = 2 → n = 4
n = 2 → n = 5
Step 3: Calculate the Energy for Each Possible He+ Transition
Transition n = 1 → n = 4:
$E_1 = 13.6 \times 2^2 \left(\dfrac{1}{1^2} - \dfrac{1}{4^2}\right)\,$eV
$E_1 = 13.6 \times 4 \left(1 - \dfrac{1}{16}\right)\,$eV
$E_1 = 54.4 \times \dfrac{15}{16}\,$eV
This value is larger than 10.2 eV, so it does not match.
Transition n = 2 → n = 3:
$E_2 = 13.6 \times 2^2 \left(\dfrac{1}{2^2} - \dfrac{1}{3^2}\right)\,$eV
$E_2 = 54.4 \left(\dfrac{1}{4} - \dfrac{1}{9}\right)\,$eV
$E_2 = 54.4 \times \dfrac{5}{36}\,$eV
This value is different from 10.2 eV.
Transition n = 2 → n = 4:
$E_3 = 13.6 \times 2^2 \left(\dfrac{1}{2^2} - \dfrac{1}{4^2}\right)\,$eV
$E_3 = 54.4 \left(\dfrac{1}{4} - \dfrac{1}{16}\right)\,$eV
$E_3 = 54.4 \times \dfrac{3}{16}\,$eV
$E_3 = 54.4 \times 0.1875\,$eV = 10.2 eV
This exactly matches the energy from the hydrogen transition.
Transition n = 2 → n = 5:
Similarly, we find that the energy is not 10.2 eV; it will be a different value.
Step 4: Conclude the Matching Transition
Among the given options, the transition from n = 2 to n = 4 in the He+ ion requires exactly 10.2 eV of energy, which is the energy of the photon emitted from the H-atom transition n = 2 → n = 1. Therefore, n = 2 → n = 4 is the correct transition.
