Chemistry Q52 – JEE-MAIN 2026

Home Chemistry Structure of AtomQuestion #616

If shortest wavelength of hydrogen atom in Lyman series is

x

, then longest wavelength in Balmer series of He

^{+}

Clue 1 of 3: Rydberg Formula

The Rydberg formula describes the wavelengths of spectral lines for hydrogen-like atoms.

🥷

Ninja Strategy•Energy-Wavelength Relationship

Recognize that lower energy transitions correspond to longer wavelengths. Compare the energy differences for the two transitions to determine if the unknown wavelength should be greater or smaller than 'x'.

🧠 Full Solution Path

Step 1: Determine 'x' for Hydrogen's Lyman Series✦ Active

For the shortest wavelength in the Lyman series of a hydrogen atom, the electron transitions from $n_{2} = \infty$ to $n_{1} = 1$ . The atomic number $Z = 1$ . Using the Rydberg formula:

\frac{1}{x} = R (1)^{2} (\frac{1}{1^{2}} - \frac{1}{\infty^{2}}) = R (1 - 0) = R

Thus, $x = \frac{1}{R}$ .

Step 2: Determine the longest wavelength for He

^{+}

's Balmer Series○ Expand

For the longest wavelength in the Balmer series of He $^{+}$ , the electron transitions from $n_{2} = 3$ to $n_{1} = 2$ . The atomic number $Z = 2$ . Let the longest wavelength be $λ^{'}$ . Using the Rydberg formula:

\frac{1}{λ^{'}} = R (2)^{2} (\frac{1}{2^{2}} - \frac{1}{3^{2}})

\frac{1}{λ^{'}} = 4 R (\frac{1}{4} - \frac{1}{9})

\frac{1}{λ^{'}} = 4 R (\frac{9 - 4}{36})

\frac{1}{λ^{'}} = 4 R (\frac{5}{36}) = R (\frac{5}{9})

💡 Teacher's Secret Hint

Remember that for He $^{+}$ , the atomic number Z is 2, and this factor is squared in the Rydberg formula.

Step 3: Relate the wavelengths○ Expand

Substitute $R = \frac{1}{x}$ from Step 1 into the equation from Step 2:

\frac{1}{λ^{'}} = \frac{1}{x} (\frac{5}{9})

λ^{'} = \frac{9 x}{5}

The longest wavelength in the Balmer series of He $^{+}$ is $\frac{9 x}{5}$ .

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Chemistry Question 52 – JEE-MAIN 2026

Clue 1 of 3: Rydberg Formula

🧠 Full Solution Path