Consider a hydrogen-like ionized atom with atomic number Z with a… - Vidyadip

MCQ

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. . . . . .

✓
$3$
B
$5$
C
$8$
D
$9$

✓

Answer

Correct option: A.

$3$

a
$\Delta E =13.6 z ^2\left[\frac{1}{ n _1^2}-\frac{1}{ n _2^2}\right]$

$E _1=13.6 z ^2\left[1-\frac{1}{4}\right]=13.6 z ^2 \times \frac{3}{4}$

$E _2=13.6 z ^2\left[\frac{1}{4}-\frac{1}{9}\right]=13.6 z^2 \times \frac{5}{36}$

$E_1-E_2=13.6 z^2\left[\frac{3}{4}-\frac{5}{36}\right]$

$\Rightarrow 13.6 z^2 \times \frac{22}{36}=74.8$

$z^2=\frac{74.8}{13.6} \times \frac{18}{11}=9$

$z=3$

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