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 to transition has energy 74.8 eV higher than the photon emitted in the to transition. The ionization en

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 $n = 2$ 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 ______. [2018]

Smart Achievers · Accepted Answer

(3)

According to the question, 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.

$∴$ $Δ E_{2 \to 1} = 74.8 + Δ E_{3 \to 2}$

$13.6 Z^{2} (1 - \frac{1}{4}) = 74.8 + 13.6 Z^{2} (\frac{1}{4} - \frac{1}{9})$

$∴$ $Z = 3$

Solution

Want Us to Email you About Special Offers & Updates?

Solution

Ans. (3) According to the question, the photon emitted in the n=2→n=1 transition has energy 74.8 eV higher than the photon emitted in the n=3→n=2 transition. ∴ ΔE2→1=74.8+ΔE3→2 13.6Z2(1-14)=74.8+13.6Z2(14-19) ∴ Z=3

Want Us to Email you About Special Offers & Updates?