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  • In a transition to the state of excitation energy 10.19 eV a hydrogen atom emits a <img alt="4890 \mathrm{~A}^0" src="https://entrancecorner.oncodecogs.com/gif.latex?4890%20%5Cmathrm%7B%7EA%7D%5E0"

In a transition to the state of excitation energy 10.19 eV a hydrogen atom emits a 4890 \mathrm{~A}^0 photon. Determine the binding energy of the initial state.

Option: 1

0.67 eV


Option: 2

0.74 eV


Option: 3

0.78 eV


Option: 4

0.87 eV


Answers (1)

best_answer

\mathrm{\text { The energy of the emitted photon is }=\mathrm{h} v=\mathrm{hc} / \lambda}

\mathrm{\text { Now, hc }=\left(6.63 \times 10^{-34} \mathrm{Js}\right) \times\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)}

\mathrm{\begin{aligned} & =19.89 \times 10^{-26} \mathrm{~J}-\mathrm{m} \\ & =19.89 \times 10^{-16} \mathrm{JA}^0 \\ & =\frac{19.89 \times 10^{-16}}{1.6 \times 10^{-19}} \mathrm{eVA} A^0=12.4 \times 10^3 \mathrm{eV} \mathrm{A}^0 \\ & \therefore \quad \text { Energy of photon }=\frac{h c}{\lambda}=\frac{12.4 \times 10^3 \mathrm{eVA}^0}{4.89 \times 10^3 A^0}=2.54 \mathrm{eV} \end{aligned}}

The excitation energy\mathrm{\mathrm{E}_{\mathrm{x}}} is the energy to excite the atom to a level above the ground state.

\mathrm{\text { Now } E_n=E_1+E_x=-13.6 \mathrm{eV}+10.19 \mathrm{eV}=-3.41 \mathrm{eV}}

Let the emission of photon occur due to the transition from a higher energy state Eh to the lower energy state

\mathrm{\begin{aligned} & E_L\left(=E_n\right) . \text { Now } E_L=-3.41 \mathrm{eV} \text { and the difference } E_h- \\ & E_L=2.54 \mathrm{eV} . \\ & \therefore \quad E_h=2.54+E_L=(2.54-3.41) \mathrm{eV}=-0.874 \mathrm{eV} . \end{aligned}}

Hence binding energy of electron in the initial state is
0.87 eV.

Posted by

Riya

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