A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980 A ~ … . The radius of the atom in the excited state, in terms of Bohr radius a 0 , will be:

Energy of photon = λ hc = 980 12500 = 12.75 eV Energy of electron in n th orbit is given by En = n 2 − 13.6 ⇒ E n − E 1 = − 13.6 [ n 2 1 1 2 − 1 ] ⇒ 12.75 = 13.6 [ 1 2 1 n 2 − 1 ] ⇒ n = 4 ∴ Electron will excite to n = 4 We know that ' R ′ ∝ n 2 ∴ Radius of atom will be 16 a 0

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PhysicsModern PhysicsAtomic structureMedium2 minPYQ_2019

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A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength $980 \tilde{A} \dots$ . The radius of the atom in the excited state, in terms of Bohr radius $a_{0}$ , will be:

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Solution:

Energy of photon $= \frac{hc}{λ} = \frac{12500}{980} = 12.75 eV$ Energy of electron in $n^{th}$ orbit is given by $En = \frac{- 13.6}{n ^{2}} \Rightarrow E_{n} - E_{1} = - 13.6 [\frac{1}{n ^{2}} \frac{- 1}{1 ^{2}}]$ $\Rightarrow 12.75 = 13.6 [\frac{1}{1 ^{2}} \frac{- 1}{n ^{2}}] \Rightarrow n = 4$ $∴$ Electron will excite to $n = 4$ We know that ' $R^{'} \propto n^{2}$ $∴$ Radius of atom will be $16 a_{0}$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: PYQ_2019

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