Question

For the similar electronic transition occurring in the following atoms or ions, the wavelength of
radiation emitted out is maximum for
(A) H-atom
(B) Het ion (C) Lil+ ion (D) Be+ ion
ni
Will​

Answer 1

answer : option (A) H - atom

explanation : from Rydberg's equation,

$\frac{1}{\lambda}=RZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

where Z is atomic number, R is Rydberg's constant and λ is wavelength of atom/ion.

a/c to question, electronic transition is similar in all the following atoms or ions.

so, $\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$ is constant.

so, $\frac{1}{\lambda}\propto Z^2$

means, wavelength is inversely proportional to atomic number of the atom.

so, wavelength will be maximum when atomic number will be minimum.

here, atomic no of H-atom , Z = 1

atomic no of He+ ion , Z = 2

atomic no of Li²+ ion , Z = 3

atomic no of Be³+ ion , Z = 4

here it is clear that atomic no of H-atom is minimum.

so, wavelength of emitted radiation is maximum for H-atom.