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Q.

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 to n = 1 transition has energy 74.8 eV higher than the photon emitted in the n = 3 to n = 2 transition. The ionization energy of the hydrogen atom is 13.6 eV. The value of Z is _________.

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answer is 3.

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Detailed Solution

As described in question, we have 
 ΔE2−1=74.8+ΔE3−2
Transition energy can be calculated as
ΔE=13.6Z2(1n12−1n22eV) 
 ⇒13.6Z2[1−14]=74.8+13.6Z2[14−19]
 ⇒Z=3

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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 to n = 1 transition has energy 74.8 eV higher than the photon emitted in the n = 3 to n = 2 transition. The ionization energy of the hydrogen atom is 13.6 eV. The value of Z is _________.