Question

(10) A positive ion having just one electron ejects it if a
photon of wavelength 228 A or less is absorbed by it.
Identify the ion.​

Answer 1

using the formula,

energy of photon in eV is given by, $E=\frac{12400}{\lambda(A^{\circ})}$

given, wavelength = 228 A°

so, energy of photon = 12400/228 eV

= 54.38 eV

now from Bohr's theory,

E = $13.6\frac{Z^2}{n^2}$eV

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 ion is $He^+$

Answer 2

A positive ion having just one electron ejects it if a  photon of wavelength 228 A or less is absorbed by it. So, helium is the ion.

Step-by-step explanation:

In the question, it is given:

Photon’s wavelength, $\lambda=228 \mathrm{A}$

Energy E is shown as $E=h c \lambda$

Here, c = light’s speed

         h = Constant of Planck

Therefore, $E=6.63 \times 10-34 \times 3 \times 10-10 \times 108228$

$=0.0872 \times 10-16 J$

When there is a transition to to n = 2 from n = 1, the excitation energy (E1) will be

$\mathrm{E} 1=13.6 \mathrm{eV} \times 112-122 \times \mathrm{Z} 2$

$\Rightarrow \mathrm{E} 1=13.6 \mathrm{eV} \times 34 \times \mathrm{Z} 2$

The photon’s energy is equal to the excitation energy.

Therefore, $z 2=5.34$

$z=5.34=2.3$

So, helium is the ion.