Question

An electron in a long, organic molecule used in a dye laser behaves approximately like a quantum particle in a box with width 4.18 nm. Find the emitted photon when the electron makes a transition from the first excited state to the ground state and from the second excited state to the first excited state.

Answer 1

The wavelength of the particle when it makes a transition from the first excited state to the ground state will be 1.92 x 10⁴ nm and when it makes a transition from the second excited state to the first excited state, its wavelength will be  1.15 x 10⁴ nm.

The wavelength of light is defined as "the distance between two successive peaks or troughs of a light wave".

• It is denoted by the Greek letter lambda (λ).

Wavelength of the emitted photon can be calculated as,

Since electron behaves like a quantum particle in a box, therefore the energy is the discrete energy and can be defined by the formula:

E = (n²$\pi$²h²) / (2*m*l)

where; n = transition state

h = planks constant = 1.054 x 10⁻²⁴

m = mass of the electron = 9.11 x 10⁻³¹ kg

l = length of the box = 4.18 x10⁻⁹ m

for the transition from the first excited state to the ground state n² = 2² -1²

Putting the value in equation of energy,

E₁ = {(2² -1²)$\pi$²(1.054 x 10⁻²⁴)²} / {2*(9.11 x 10⁻³¹)*(4.18 x10⁻⁹)}

E₁ = 1.03 x 10⁻²⁰ joules

E₁ = 1.03 x 10⁻²⁰ / 1.6 x 10⁻¹⁹ ev

E₁ = 6.46 x 10⁻² ev

As we know that ΔE = (hc/λ)

therefore λ = (hc/ΔE )

ΔE = E₁

λ = 1240 / 6.46 x 10⁻²

λ = 1.92 x 10⁴ nm

Similarly, calculating energy for the the transition of the photon from the second excited state to the first excited state came out to be

E₂ = 1.72 x 10⁻²⁰ joules

E₂ = 0.108 ev

And so,

λ =  1.15 x 10⁴ nm

The wavelength of the particle will be 1.92 x 10⁴ nm and 1.15 x 10⁴ nm for its transition from first to ground and second to first excited state respectively.

To know more about wavelength,

https://brainly.com/question/13884194

#SPJ4