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JEE Main Chemistry Structure of Atom 2026 JEE Main 2026 (24 January Shift 1)

JEE Main Chemistry Question (2026) — Solution

Question

The hydrogen spectrum consists of several spectral lines in Lyman series ( L _ 1 , ~L _ 2 , L _ 3 ; L _ 1 has lowest energy among Lyman series). Similarly it consists of several spectral lines in Balmer series ( B _ 1 , ~B _ 2 , ~B _ 3 ; B _ 1 . has lowest energy among Balmer lines). The energy of L_ 1 is x times the energy of B_ 1 . The value of x is \_\_\_\_ 10^ -1 . (Nearest integer)

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

The energy of a spectral line in the hydrogen spectrum is given by E = 13.6 ( 1 n_1^2 - 1 n_2^2 ) eV. For the Lyman series, the lowest energy line L_1 corresponds to the transition from n=2 to n=1. E(L_1) = 13.6 ( 1 1^2 - 1 2^2 ) = 13.6 ( 1 - 1 4 ) = 13.6 3 4 eV. For the Balmer series, the lowest energy line B_1 corresponds to the transition from n=3 to n=2. E(B_1) = 13.6 ( 1 2^2 - 1 3^2 ) = 13.6 ( 1 4 - 1 9 ) = 13.6 5 36 eV. Given E(L_1) = x E(B_1), we have: 13.6 3 4 = x 13.6 5 36 3 4 = x 5 36 x = 3 4 36 5 = 3 9 5 = 27 5 = 5.4 The value of x is 5.4 = 54 10^ -1 . Comparing with the required format x = integer 10^ -1 , the integer is 54.

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