Separation energy of a hydrogen like ion from its third excited state is 2.25 times the separation energy of hydrogen atom from its first excited state. Find out the atomic number of this hydrogen like ion.

Question

Separation energy of a hydrogen like ion from its third excited state is 2.25 times the separation energy of hydrogen atom from its first excited state. Find out the atomic number of this hydrogen like ion. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Let the separation energy of the ion be E . Then E = 13.6× Z2× (1/16) eV and EH=13.6× (1/4) eV â (E/EH)=Z2× (4/16) â (2.25 EH/EH)=(Z2/4) ⇒ Z2=9⇒ Z=3

Q. Separation energy of a hydrogen like ion from its third excited state is $2.25$ times the separation energy of hydrogen atom from its first excited state. Find out the atomic number of this hydrogen like ion.

Let the separation energy of the ion be $E$ . Then
$E = 13.6 \times Z^{2} \times \frac{1}{16} e V$
and $E_{H} = 13.6 \times \frac{1}{4} e V$
⇒ $\frac{E}{E _{H}} = Z^{2} \times \frac{4}{16}$
⇒ $\frac{2.25 E _{H}}{E _{H}} = \frac{Z ^{2}}{4}$
$\Rightarrow Z^{2} = 9 \Rightarrow Z = 3$

Q. Separation energy of a hydrogen like ion from its third excited state is 2.25 times the separation energy of hydrogen atom from its first excited state. Find out the atomic number of this hydrogen like ion.

Let the separation energy of the ion be E . Then E=13.6×Z2×161​eV and EH​=13.6×41​eV ⇒ EH​E​=Z2×164​ ⇒ EH​2.25EH​​=4Z2​ ⇒Z2=9⇒Z=3

Q. Separation energy of a hydrogen like ion from its third excited state is $2.25$ times the separation energy of hydrogen atom from its first excited state. Find out the atomic number of this hydrogen like ion.

Let the separation energy of the ion be $E$ . Then
$E = 13.6 \times Z^{2} \times \frac{1}{16} e V$
and $E_{H} = 13.6 \times \frac{1}{4} e V$
⇒ $\frac{E}{E _{H}} = Z^{2} \times \frac{4}{16}$
⇒ $\frac{2.25 E _{H}}{E _{H}} = \frac{Z ^{2}}{4}$
$\Rightarrow Z^{2} = 9 \Rightarrow Z = 3$