Question

An electron is in an infinite square well of width 9.6 nm. the electron makes the transition from the n =14 to the n = 11 state. calculate the wavelength of the emitted photon.

Answer 1

The wavelength of the emitted photon is 4048.5 nm.

Explanation:

Given that,

Width = 9.6 nm

Initial state =14

Final state =11

We need to calculate the wavelength of the emitted photon

Using formula of energy levels

$E_{photon}=\dfrac{n^2\pi^2h^2}{2ml^2}$

Put the value into the formula

$E_{photon}=\dfrac{14^2\pi^2\times h^2}{2ml^2}-\dfrac{11^2\pi^2\times h^2}{2ml^2}$

$E_{photon}=\dfrac{\pi^2h^2}{2ml^2}(14^2-11^2)$

$E_{photon}=\dfrac{75\pi^2h^2}{2ml^2}$

Using relation the energy of a photon

$E_{photon}=\dfrac{2\pi hc}{\lambda}$

Put the value of energy of photon

$\dfrac{75\pi^2h^2}{2ml^2}=\dfrac{2\pi hc}{\lambda}$

$\lambda=\dfrac{4cml^2}{75\pi\times h}$

Put the value into the formula

$\lambda=\dfrac{4\times3\times10^{8}\times9.1\times10^{-31}\times(9.6\times10^{-9})^2}{75\times\pi\times1.055\times10^{-34}}$

$\lambda=4048.5\ nm$

Hence, The wavelength of the emitted photon is 4048.5 nm.

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Topic : energy of photon

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