The longest wavelength of Lyman series for hydrogen atom is the same as the wavelength of certain line in spectrum of He+, when the electron makes a translation from n → 2. What is the value of n ?

The longest wavelength of the Lyman series for a hydrogen atom corresponds to the smallest energy transition, which is from the n=2 to n=1 energy level.

For the helium ion He+, the energy levels are similar to hydrogen but are scaled by a factor of Z^2, where Z is the atomic number. For helium, Z=2, so the energy levels are four times those of hydrogen.

Therefore, the energy transition corresponding to the longest wavelength in the Lyman series for He+ is four times larger than the smallest energy transition in hydrogen, which is from n=2 to n=1.

To find the initial energy level n for the transition in He+, we can set up the following equation using the Rydberg formula for the energy levels of hydrogen and helium:

1/λ = R_H * (1/2^2 - 1/n^2) = Z^2 * R_H * (1/1^2 - 1/2^2)

Solving for n, we get n = 4.

Therefore, the initial energy level for the transition in He+ that corresponds to the longest wavelength in the Lyman series is n=4.

The Lyman series in the hydrogen atom corresponds to electronic transitions that terminate in the n=1 orbit. The longest wavelength (which corresponds to the lowest frequency and energy) in the Lyman series corresponds to the n=2 to n=1 transition.

The Rydberg formula for the wavelength λ of light emitted in a hydrogen atom transition is:

1/λ = R_H * (1/n1² - 1/n2²)

where R_H is the Rydberg constant for hydrogen, n1 is the lower energy level, and n2 is the higher energy level. For the longest wavelength in the Lyman series, n1=1 and n2=2.

Now, for the helium ion He+, the same formula applies, but with a different Rydberg constant R_He+. The transition is from n to 2, so n1=2 and n2=n.

Setting the two expressions for 1/λ equal to each other gives:

R_H * (1/1² - 1/2²) = R_He+ * (1/2² - 1/n²)

We know that R_He+ = 4*R_H (because the Rydberg constant is proportional to the square of the atomic number Z, and for He+, Z=2), so we can substitute that in:

R_H * (1 - 1/4) = 4*R_H * (1/4 - 1/n²)

Solving this equation for n gives n = 3. So the transition in He+ that produces a line with the same wavelength as the longest-wavelength line in the Lyman series of hydrogen is the n=3 to n=2 transition.