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Q12P

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Fundamentals Of Physics
Found in: Page 1215

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Short Answer

An electron is trapped in a one-dimensional infinite well of width250pm and is in its ground state. What are the (a) longest, (b) second longest, and (c) third longest wavelengths of light that can excite the electron from the ground state via a, single photon absorption?

  1. The longest wavelength is 68.6 nm .
  2. The second longest wavelength is 25.72 nm .
  3. The third longest wavelength is 13.72 nm .
See the step by step solution

Step by Step Solution

Step 1: Introduction

An electron is a negatively charged subatomic particle. It can be either free (not attached to any atom), or bound to the nucleus of an atom. Electrons in atoms exist in spherical shells of various radii, representing energy levels. The larger the spherical shell, the higher the energy contained in the electron.

Step 2: Concept

An infinite potential well is a device for confining an electron. From the confinement principle we expect that the matter wave representing a trapped electron can exist only in a set of discrete states.

Energy of electron in an energy level in one-dimensional potential well is,

En=n2h28mL2

Here, h is the Planck’s constant, L is the length of the well, and m is the mass of the electron, and n is the integer.

Planck’s constant, h=6.626×10-34 Js

Mass of the electron, m=9.109×10-31 kg

Convert the length of the well from Pico meters to meters as follows:

L=250 pm =250 pm10-12 m/pm =250×1012 m

Step 3: Calculation

When electron jumps from n=n1 state to n=nf state by absorbing a

Photon, then the frequency of the photon is given by

f=Eh =h28mL21hnf2-ni2

Wavelength of the photon, λ=cf

Substitute h28mL21hnf2-ni2 for f in the above equation.

λ=ch28mL2nf2-ni2 =8mL2hnf2-ni2

Here, the electron is the ground state is ni=1

Hence, the wavelength is

λ=8mL2hnf2-ni2

Substitute 9.10×10-31 kg for m, 250×10-12 m for L, 2.998×108 m/s for c , 6.626×10-34 j.s for h , and 1 for ni in the above equation.

λ=89.10×10-31 kg250×10-12 m22.998×108 m/s6.626×10-34 j.snf2-12=2.058×107 mnf2-1=2.058×10-9 m109 nm/mnf2-1=205.8 nmnf2-1

Step 4: (a) Determine the longest wavelength

For longest wavelength, nf=2

The longest wavelength is,

λ=205.8 nmnf2-1

Substitute 2 for in the above equation.

role="math" localid="1661859531615" λ=205.8 nm22-1 =68.6 nm

Therefore, the longest wavelength is 68.6 nm .

Step 5: (b) Determine the second longest wavelength.

For second longest wavelength, nf=3

The second longest wavelength is,

λ=205.8 nmnf2-1

Substitute 3 for nf in the above equation.

λ=205.8 nm32-1 =25.72 nm

Therefore, the second longest wavelength is 25.72 nm.

Step 6: (c) Determine third longest wavelength

For third longest wavelength, nf=4

The third longest wavelength is,

λ=205.8 nmnf2-1

Substitute 4 for nf in the above equation.

λ=205.8 nm42-1 =13.72 nm

Therefore, the third longest wavelength is 13.72 nm.

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