An electron in a hydrogen-like atom jumps from the 3^ rd energy state to the ground state. If the energy released in this process is 108.8 eV , the atomic number of the atom is: (Given Rch = 13.6 eV )

Step-by-step solution

The energy released when an electron jumps from a higher energy state n₂ to a lower energy state n₁ in a hydrogen-like atom is given by: E = 13.6 Z^2 ( 1 n₁^2 - 1 n₂^2 ) eV Here, the electron jumps from the 3^ rd energy state ( n₂ = 3 ) to the ground state ( n₁ = 1 ). The energy released is given as 108.8 eV . Substituting the given values into the formula: 108.8 = 13.6 Z^2 ( 1 1^2 - 1 3^2 ) 108.8 = 13.6 Z^2 ( 1 - 1 9 ) 108.8 = 13.6 Z^2 ( 8 9 ) 108.8 13.6 = Z^2 ( 8 9 ) 8 = Z^2 ( 8 9 ) Z^2 = 9 Z = 3 The atomic numbe