If on transition to the ground state an H e + ion emits two photons in succession, having wavelengths 1026 . 7 A ∘ and 304 A ∘ , then the quantum number n corresponding to the exciting state of H e + ion is [ R = 1 .096 × 10 7 m - 1 ]

h c λ 1 + h c λ 2 = R c h Z 2 1 1 2 - 1 n 2 Put λ 1 = 1026.7 Å and λ 2 = 304 Å Z = 2 for H e + ion On solving for n n = 6

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PhysicsModern PhysicsAtomic structureEasy2 minPYQ_2020

PhysicsEasysingle choice

If on transition to the ground state an $H e^{+}$ ion emits two photons in succession, having wavelengths $1026.7 \overset{\circ}{A}$ and $304 \overset{\circ}{A}$ , then the quantum number $n$ corresponding to the exciting state of $H e^{+}$ ion is [ $R = 1 .096 \times 10^{7} m^{- 1}$ ]

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Answer:

Solution:

$\frac{h c}{λ_{1}} + \frac{h c}{λ_{2}} = R c h Z^{2} (\frac{1}{1^{2}} - \frac{1}{n^{2}})$

Put $λ_{1} = 1026.7 Å$ and $λ_{2} = 304 Å$

$Z = 2$ for $H e^{+}$ ion

On solving for $n$

$n = 6$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: PYQ_2020

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