Answer:
[tex]\mathbf{\lambda \simeq 3.039 \times 10^{-8} \ m}[/tex]
Explanation:
For a hydrogen-like atom, the spectral line wavelength can be computed by using the formula:
[tex]\bar v = Z^2 R_H \Big(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}\Big)[/tex]
where:
emitted radiation of the wavenumber [tex]\bar v[/tex] = ???
atomic no of helium Z = 2
Rydberg's constant [tex]R_H = 1.097*10^7 \ m^{-1}[/tex]
the initial energy of the principal quantum [tex]n_1[/tex] = 2
the initial energy of the principal quantum [tex]n_1[/tex] = 2
Now, the emitted radiation of the wavenumber can be computed as:
[tex]\bar v = (2)^2 (1.097*10^7 \ m^{-1} ) \Big(\dfrac{1}{1^2}-\dfrac{1}{2^2}\Big)[/tex]
[tex]\bar v = 3.291 \times 10^ 7/m[/tex]
Now, the wavelength for the transition can be computed by using the relation between the wavelength λ and the emitted radiation of the wavenumber [tex]\bar v[/tex], which is:
[tex]\bar v = \dfrac{1}{\lambda}[/tex]
[tex]\lambda = \dfrac{1}{\bar v}[/tex]
[tex]\lambda = \dfrac{1}{3.291 \times 10^{7}}\times \dfrac{m}{1}[/tex]
[tex]\mathbf{\lambda =3.03859 \times 10^{-8} \ m}[/tex]
[tex]\mathbf{\lambda \simeq 3.039 \times 10^{-8} \ m}[/tex]
