The frequency of the 1^{st} line of the Balme | vedclass

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Question

(B) The frequency of a spectral line in a hydrogen-like atom is given by the Rydberg formula: $
u = R c Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2}

Vedclass · Accepted Answer

$4{
u _0}$

Vedclass · Answer

(B) The frequency of a spectral line in a hydrogen-like atom is given by the Rydberg formula: $
u = R c Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the $1^{st}$ line of the Balmer series,$n_1 = 2$ and $n_2 = 3$.Thus,$
u \propto Z^2$.For hydrogen $({H_2})$,$Z = 1$,so $
u_{H} = k(1)^2 = 
u_0$.For singly ionized helium $({He^+})$,$Z = 2$,so $
u_{He} = k(2)^2 = 4k = 4
u_0$.Therefore,the frequency of the corresponding line for ${He^+}$ is $4
u_0$.

The frequency of the $1^{st}$ line of the Balmer series in the ${H_2}$ atom is ${\nu _0}$. What is the frequency of the corresponding line emitted by a singly ionized ${He^+}$ atom?

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