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Bohr's Model of an Atom

He atom can b...

Question

He atom can be excited to $I s^{1} 2 p^{1}$ by $λ = 58.44 n m$ . If lowest excited state for $H e$ lies $4857 c m^{- 1}$ below the above. Calculate the energy of the lower excitation state.

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Solution

$E_{1} = \frac{h c}{λ_{1}} = \frac{1242}{58.44} = 21.25 e V$
For lowest excitation state $λ_{2} = \frac{1}{485700} = 2.05 \times 10^{- 6} m = 2050 n m E_{2} = \frac{1242}{2050} = 0.606 e V$
If lowest excitation state lied $0.626 e V$ below the $21.25 e V$
Energy of lowest excitation state= $21.25 - 0.606 = 20.644 e V$

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