A triply ionized beryllium atom is in the ground state. It absorbs energy and ma

Question

A triply ionized beryllium atom is in the ground state. It absorbs energy and makes a transition to the n = 3 excited state. The ion returns to the ground state by emitting two photons. What are their wavelengths?
1 m (lower energy photon)
2 m (higher energy photon)

Explanation / Answer

The transition has to go n = 3 -> n=2 then n=2 ->n=1 The energy of the photons is E32 = -13.6 eV*(1/3^2 - 1/2^2) = 1.889 eV E21 = -13.6eV(1/2^1 -1/1) = 10.2 eV Now E = hc/l for a photon where h= planck's constant, c = speed of light and l = wavelength 1 eV = 1.6 x 10^-19 J So l32 = hc/E32 = 6.62x10^034*3x10^8/(1.6x10^-19*1.889) = 0.657 x 10^-6 m l21 = hc/E21 = 0.122x10^-6 m

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