Question

The allowed energies of a quantum system are 0.0 eV, 5.0 eV , and 8.5 eV .

Answer 1

Explanation:

Given that,

Energies of a quantum system

E₁ = 0.0 eV

E₂ = 5.0 eV

E₃ = 8.5 eV

We need to calculate the wavelength

Using formula of energy

$E=\dfrac{hc}{\lambda}$

An electron transits from 5.0 eV to 0.0 eV

Thereby emitting a photon

For E₁ = 5.0 eV,

Using formula of energy

$E_{1}=\dfrac{hc}{\lambda_{1}}$

Where, E = energy

h = Planck constant

$\lambda$ = wavelength

Put the value into the formula

$5.0\times1.6\times10^{-19}=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{\lambda_{1}}$

$\lambda_{1}=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{5.0\times1.6\times10^{-19}}$

$\lambda_{1}=2.486\times10^{-7}\ m$

$\lambda_{1}=248.6\ nm$

An electron transits from 8.5 eV to 0.0 eV

Thereby emitting a photon

For E₂ = 8.5 eV,

The wavelength of photon

$\lambda_{2}=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{8.5\times1.6\times10^{-19}}$

$\lambda_{2}=1.463\times10^{-7}\ m$

$\lambda_{2}=146.3\ nm$

An electron transits from 8.5 eV to 5.0 eV

Thereby emitting a photon

For E₃ = 8.5-5.0 =3.5 eV,

The wavelength of photon

$\lambda_{3}=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{3.5\times1.6\times10^{-19}}$

$\lambda_{3}=3.5517\times10^{-7}$

$\lambda_{3}=355.2\ nm$

Hence, This is the required solution.