Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = 2   t o   n =   1 transition has energy 74.8   eV higher than the photon emitted in the n = 3   t o   n = 2…

Question

Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = 2 &#160; t o &#160; n = &#160; 1 transition has energy 74.8 &#160; eV higher than the photon emitted in the n = 3 &#160; t o &#160; n = 2 transition. Given that the ionization energy of the hydrogen atom is 13.6 &#160; eV , what is the value of Z ?

Aicharya · Accepted Answer

∆ E 2 &#x2192; 1 = 13.6 × Z 2 1 - 1 4 = 13.6 × Z 2 3 4 ∆ E 3 &#x2192; 2 = 13.6 × Z 2 1 4 - 1 9 = 13.6 × Z 2 5 36 ∆ E 2 &#x2192; 1 = ∆ E 3 - 2 + 74.8 13.6 × Z 2 3 4 = 13.6 × Z 2 5 36 + 74.8 13.6 × Z 2 3 4 - 5 36 = 74.8 Z 2 = 9 Z = + 3 `

Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum…

Solution