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PhysicsModern PhysicsAtomic structureEasy2 minPYQ_2019

PhysicsEasysingle choice

An excited ${H e}^{+}$ ion emits two photons in succession, with wavelengths $108.5 n m$ and $30.4 n m$ in making a transition to the ground state. The quantum number $n$ , corresponding to its initial excited state is

(for a photon of wavelength $λ$ , energy $E = \frac{1240 eV}{λ (in nm)}$ )

Options:

Answer:

Solution:

$E_{1} = \frac{1240}{λ_{1}} = \frac{1240}{108.5} \approx 11.43 e V$
$E_{2} = \frac{1240}{λ_{2}} = \frac{1240}{30.4} \approx 40.79 e V$
$E_{T o t a l} = E_{1} + E_{1} = 52.22 e V$
$52.22 = 13.6 {(2)}^{2} [1 - \frac{1}{n^{2}}]$
$52.22 = 54.4 (1 - \frac{1}{n^{2}})$
$0.96 = 1 - \frac{1}{n^{2}}$
$\frac{1}{n^{2}} = 0.04$
$n^{2} = \frac{1}{0.04} = \frac{100}{4} = 25$
$n = 5$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: PYQ_2019

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