Finf the wavelength of the first line of the `HE^(+)`ion spectral series whose interval between the extreme line is `Delta omega= 5.18.10^(15) s^(-1)`.

Finf the wavelength of the first line of the `HE^(+)`ion spectral seri

1.	Finf the wavelength of the first line of the `HE^(+)`ion spectral series whose interval between the extreme line is `Delta omega= 5.18.10^(15) s^(-1)`.
Answer» We start from the generalized Balmer formula `omega=RZ^(2)((1)/n^(2)-(1)/(m^(2)))` Here `m=n+1, n+2,....oo` Then the angular frequency of the first line of this series (series `n`) is `omega_(1)=RZ^(2)((1)/(n^(2))-(1)/((n+1)^(2)))=Delta omega(((n+1)/(n))^(2)-1)` `=Delta omega [{(Zsqrt((R)/(Delta omega)))/(Zsqrt((R)/(Deltaomega)))}-1]` Then the wavelength will be `lambda_(1)=(2pi c)/(omega_(1))=(2pi c)/(Delta omega)((Zsqrt((R)/(Delta omega)-1)))/(2Zsqrt((R)/(Delta omega))-1)` Substitution (with the value `R` from problem 6.34 which is also the correct value determined directly) gives `lambda_(1)= 0.468mu m`.

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Finf the wavelength of the first line of the `HE^(+)`ion spectral series whose interval between the extreme line is `Delta omega= 5.18.10^(15) s^(-1)`.

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