Shortest wavelength in lyman setirs= 1/R which is equal to X
1/R=X
longest wavelength in balmer series of helium ion:
n1=2, n2=3, Z=2
1/wavelength = R(1/4 - 1/9)×2^2
=R(5/36)×4
wavelength= 9/5R
= 9X/5

Calculation of Longest Wavelength in Balmer Series of Helium Ion:
The Balmer series for hydrogen atom is a set of spectral lines corresponding to transitions from higher energy levels to the second energy level. The longest wavelength in the Balmer series of hydrogen atom is 656.3 nm.

Relationship between Lyman and Balmer Series:
For hydrogen atom, the Lyman series corresponds to transitions to the first energy level. The shortest wavelength in the Lyman series is X.

Wavelength Relationship between Hydrogen and Helium Ion:
For helium ion (He+), the energy levels and transitions are different compared to hydrogen atom. However, we can use the relationships between the wavelengths of hydrogen and helium ions to determine the longest wavelength in the Balmer series of helium ion.

Calculating the Longest Wavelength in Balmer Series of Helium Ion:
The formula to calculate the wavelength of a spectral line is given by:
1/λ = R(1/n1^2 - 1/n2^2)
where R is the Rydberg constant, n1 is the initial energy level, and n2 is the final energy level.
For the Balmer series of helium ion (He+), the initial energy level is 3 (corresponding to the second energy level) and the final energy level can vary. To find the longest wavelength, we need to consider the transition with the highest energy difference, which corresponds to the transition to the fourth energy level (n=4).
Plugging in the values, we get:
1/λ = R(1/3^2 - 1/4^2)
Solving the equation will give us the longest wavelength in the Balmer series of helium ion.
By following this approach, you can calculate the longest wavelength in the Balmer series of helium ion based on the given information about the Lyman series of hydrogen atom.