Verified answer

[tex]\huge\pink{\mid{\fbox{\tt{ᴀɴsᴡᴇʀ }}\mid}}[/tex]

[tex]E=\frac{12400}{\lambda(A^{\circ})}[/tex][/tex]

given, wavelength = 228 A°

so, energy of photon = 12400/228 eV

= 54.38 eV

now from Bohr's theory,

[tex]E =

13.6\frac{Z^2}{n^2}[/tex]

where Z is atomic number and n is nth orbital from which an electron is ejected.

so, 54.38 eV = 13.6 × Z²/12

or, Z² = (54.38/13.6) ≈ 4

or, Z = 2

hence, atomic number of positive ion is 2. so, element is not other than Helium.

and

[tex]is \: \: ion \: \: He^+[/tex]