Question

The wave number of photon emitted when electron in 4th excited state make transition to 2nd orbit in He+ ion is (R = 1.09 × 105 cm–1)


9.16 × 104 cm–1


8.4 × 106 cm–1


1.06 × 105 cm–1


7.43 × 102 cm–1

Answer 1

Given:

Initial excited state, n1 = 4

Final excited state, n2 = 2

R = 1.09 × 10⁵ cm⁻¹

To Find:

The wavenumber of the photon emitted.

Calculation:

- For He⁺, Z = 2

- Wave number is given as:

Wavenumber = R {1/n1² - 1/n2²} × Z²

⇒ Wavenumber = 1.09 × 10⁵ {1/2² - 1/4²} × 2²

⇒ Wavenumber = 1.09 × 10⁵ {1/4 - 1/16} × 4

⇒ Wavenumber = 1.09 × 10⁵ × 3/16 × 4

⇒ Wavenumber = 0.8175 × 10⁵

⇒ Wavenumber = 8.175 × 10⁴ cm⁻¹

- So, none of the given options is correct.

The correct answer for wavenumber is  8.175 × 10⁴ cm⁻¹.