The energy required to ionise a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state ?

Question

The energy required to ionise a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state ? (A) 8.6 nm (B) 24.2 nm (C) 11.4 nm (D) 35.8 nm

Tardigrade Admin · Accepted Answer

beginarrayl E n =-( text Ionisation energy / n 2) =(- q R / n 2) endarray now, when it jumps, the released energy will be beginarrayl E 2- E 1=- qR ((1/9)-1) =8 R endarray beginarrayl text Now, E =( hc /λ)=8 R λ=( hc /8 R ) =(6.64 × 10-34 × 3 × 108/8 × 2.2 × 10-18) =11.31 nm ≈ 11.4 nm endarray

Q. The energy required to ionise a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state ?

$8.6 nm$

$24.2 nm$

$11.4 nm$

$35.8 nm$

Q. The energy required to ionise a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state ?

8.6nm

24.2nm

11.4nm

35.8nm

En​=−n2 Ionisation energy ​=n2−qR​​ now, when it jumps, the released energy will be E2​−E1​=−qR(91​−1)=8R​ Now, E=λhc​=8Rλ=8Rhc​=8×2.2×10−186.64×10−34×3×108​=11.31nm≈11.4nm​

$8.6 nm$

$24.2 nm$

$11.4 nm$

$35.8 nm$