In a single isolated atom of He +ion an electron makes transition from third excited state to ground state by emitting only two photons. The higher energy (in eV ) will be:

Question

In a single isolated atom of He +ion an electron makes transition from third excited state to ground state by emitting only two photons. The higher energy (in eV ) will be: (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/43e5411b0969ab0c303913c29f3b699e-.png" /> Since electron is coming from 3rd excited state to ground state by emitting only two photrons, it can be 4 arrow 3 and 3 arrow 1 Hence, higher energy value will be Δ E3 arrow 1=(13.6-1.5) × 22=48.4 eV

Q. In a single isolated atom of $H e^{+}$ ion an electron makes transition from third excited state to ground state by emitting only two photons. The higher energy (in $e V$ ) will be:

Since electron is coming from $3 r d$ excited state to ground state by emitting only two photrons, it can be $4 \to 3$ and $3 \to 1$
Hence, higher energy value will be
$Δ E_{3 \to 1} = (13.6 - 1.5) \times 2^{2} = 48.4 e V$

Q. In a single isolated atom of He+ion an electron makes transition from third excited state to ground state by emitting only two photons. The higher energy (in eV ) will be:

Since electron is coming from 3rd excited state to ground state by emitting only two photrons, it can be 4→3 and 3→1 Hence, higher energy value will be ΔE3→1​=(13.6−1.5)×22=48.4eV

Q. In a single isolated atom of $H e^{+}$ ion an electron makes transition from third excited state to ground state by emitting only two photons. The higher energy (in $e V$ ) will be:

Since electron is coming from $3 r d$ excited state to ground state by emitting only two photrons, it can be $4 \to 3$ and $3 \to 1$
Hence, higher energy value will be
$Δ E_{3 \to 1} = (13.6 - 1.5) \times 2^{2} = 48.4 e V$