Question

Consider an electron in an infinite well of width 0.7 nm . What is the wavelength of a photon emitted when the electron in the infinite well makes a transition from the first excited state to the ground state? The value of ¯h is 1.05457 × 10−34 J · s, the Bohr radius is 5.29177 × 10−11 m , the Rydberg constant for hydrogen is 1.09735 × 107 m−1 , the ground state energy for hydrogen is 13.6057 eV , and the speed of light is 2.99792 × 108 m/s. 1 eV = 1.60218 × 10−19 eV . Answer in units of nm.

Answer 1

λ = 538.0 nm

Step-by-step explanation:

The solution of the Schrödinger equation for the inner part of the well gives energy

`E_(n)` = (h² / 8mL²) n²

Where n is an integer and L is the length of the well

They ask for the transition from the first excited state n = 2 to the base state n = 1

E₂ - E₁ = = (h² / 8mL²) (n₂² - n₁²)

Let's calculate

E₂-E₁ = (6.63 10⁻³⁴)² / (8 9.1 10⁻³¹ (0.7 10⁻⁹)²) (2² -1²)

E₂ –E₁ = 3.6968 10⁻¹⁹ J

Let's use the Planck equation

E = h f

c = λ f

E = h c / λ

E = E₂ ₂- E₁

h c / λ = 3.6968 10⁻¹⁹

λ = h c / (E₂-E₁)

λ = 6.63 10⁻³⁴ 3 10⁸ / 3.6968 10⁻¹⁹

λ = 5.380 10⁻⁷ m

Let's reduce

λ = 5.380 10⁻⁷ m (10 9 nm / 1 m)

λ = 538.0 nm