Q9P - Fundamentals Of Physics

Question

Suppose that an electron trapped in a one-dimensional infinite well of width $\mathbf{250}\mathit{p}\mathit{m}$  is excited from its first excited state to its third excited state. 

(a) What energy must be transferred to the electron for this quantum jump? The electron then de-excites back to its ground state by emitting light. In the various possible ways it can do this, what are the (b) shortest, (c) second shortest, (d) longest, and (e) second longest wavelengths that can be emitted? (f) Show the various possible ways on an energy-level diagram. If the light of wavelength   $\mathbf{29}\mathbf{.}\mathbf{4}\mathit{n}\mathit{m}$happens to be emitted, what are the (g) longest and (h) shortest wavelength that can be emitted afterward

Step-by-Step Solution

Answer

(a) The energy required to excite an electron from its first excited state to its third excited state is $72.2eV$ .

 

(b) The shortest wave length that can be is $13.7n\mathrm{m}$ .

 

(c) The shortest wave length that can be is  $17.2n\mathrm{m}$.

 

(d) The shortest wave length that can be is $68.7n\mathrm{m}$ .

 

(e) The shortest wave length that can be is  $41.2n\mathrm{m}$.

 

(f) The possible downwards transition of an electron from the third excited state to the

      ground state.

 

(g) The emitted wavelength 29.4 nm corresponds to the transition from third excited state to the second excited state  $4\to 3$.

 

(h) The shortest wavelength that is  $25.8n\mathrm{m}$.

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 that represent energy levels. The larger the spherical shell, the higher the energy contained in the electron.

Step 2: Concept:

Expression for the energy required to excite an electron from a lower energy state quantum number to higher energy state quantum number in a one-dimensional potential well is 

  $∆E=\frac{h^2}{8m{L}^2}({n^2}_{high}-{n^2}_{Low})$   

 

Here,   $∆E$is the energy required to excite electrons from low energy state principal

Quantum number $n_{low}$  to higher energy state principal quantum number $n_{high}$  trapped in one dimensional infinite potential well,  $h$ is the Planck’s constant,  $m$ is the mass of electron,   $L$is the width of the potential well.

 

Quantum number  $n$ takes positive integer values$n=1,2,3,\dots \dots ,$  

 

Here,  $n=1$ represents zero point energy state, ground state, $n=2$  represents the second excited state and $n=3$  represents the second excited state and so on.

Step 3: (a) Find the energy required to excite electron

Rewrite equation (1) and multiply and divide by   $c^2$ 

 $∆E=\frac{{\left(hc\right)}^2}{8{\left(mc\right)}^2{L}^2}({n^2}_{high}-{n^2}_{Low})$

Here, c is the speed of light.

 

Subscribe   $1240ev.nm$ for  $hc$,  $0.511MeV$ for $m{c}^2$ , 4 for $n_{high}$ ,2 for  $n_{low}$ , and   $250pm$for  $L$ to find  $∆E$.

$$\begin{gathered} ∆E=\frac{{\left(1240ev.nm\right)}^2}{8\times 0.511MeV\times \left({\displaystyle \frac{{10}^{6}ev}{1MeV}}\right)\times {\left(250pm\times \left({\displaystyle \frac{{10}^{-3}nm}{1pm}}\right)\right)}^2}\left({\left(4\right)}^2-{\left(2\right)}^2\right) \\ =72.2eV \end{gathered}$$ 

 

Therefore, the energy required to excite electron from its first excited state to its

Third excited state is $72.2eV$ .

 

Starting from third excited state  $n=4$, the electron can reach the ground state$n=1$   by either making a down ward quantum jump directly to the ground state energy level or by making separate jumps all the way to the ground state by emitting a photon with energy equal to the energy difference between the energy levels. There are 8 possible transition of electron from third excited state to ground state.

 $$\begin{gathered} n=4\to 1 \\ n=4\to 2then n=2\to 1 \\ n=4\to 3then n=3\to 1 \\ n=4\to 3then n=3\to 2,then3\to 1 \end{gathered}$$

 

Express the relation for the wave length of the emitted photon 

${\lambda }_{high\to low}=\frac{8\left(m{c}^2\right)L^2}{hc\left({n^2}_{high}-{n^2}_{low}\right)}$                                                                              ….. (2)

 

Here,  $\lambda$ is the wave length of the emitted photon when the electron jumps from  $n_{high}$ quantum state to  $n_{low}$ quantum state .

 

The spacing between the energy level gets reduced as we move up from ground state to higher quantum state. Refer fig 3.18 (a) the expression between the energy and the wave length is,

 $∆E=\frac{hc}{∆\lambda }$

The wave length is inversely proportional to the difference in energy.

 

To have largest wave length emitted then,

The difference in the energy level must be smallest or vice versa.

Step 4: (b) Find the shortest wavelength:

For the shortest wavelength the electron has to move down from $n_{high}=4$   to   $n_{low}=1$quantum state.

Rewrite equation (2) as below.

 ${\lambda }_{high\to low}=\frac{8\left(m{c}^2\right)L^2}{hc\left({n^2}_{high}-{n^2}_{low}\right)}$

 

Substitute  $1240ev.nm for hc, 0.511MeV for m{c}^2, 4 for n_{high}, 1 for n_{low} and for 250pm for L$ in the above equation. 

 

 $$\begin{gathered} {\lambda }_{4\to 1}=\frac{8\times 0.511MeV\times \left({\displaystyle \frac{{10}^{6}ev}{1MeV}}\right)\times {\left(250pm\times \left({\displaystyle \frac{{10}^{-3}nm}{1pm}}\right)\right)}^2}{\left(1240eV.nm\right)\times \left({\left(4\right)}^2-{\left(1\right)}^2\right)} \\ =\frac{2,55,500eV.n{m}^2}{\left(1240ev.nm\right)\times 15} \\ =13.7nm \end{gathered}$$

Therefore, the shortest wave length that can be is  $13.7nm$.

Step 5: (c) Find the second shortest wavelength:

The second shortest wave length emitted when electron moves down from   $n_{high}=4$to $n_{low}=2$  quantum state.

 

Rewrite equation (2) as below.

 ${\lambda }_{high\to low}=\frac{8\left(m{c}^2\right)L^2}{hc\left({n^2}_{high}-{n^2}_{low}\right)}$

 

Substitute  $1240ev.nm for hc, 0.511MeV for m{c}^2, 4 for n_{high}, 2for n_{low} and for 250pm for L$   in the above expression. 

 

 $$\begin{gathered} {\lambda }_{4\to 2}=\frac{8\times 0.511MeV\times \left({\displaystyle \frac{{10}^{6}ev}{1MeV}}\right)\times {\left(250pm\times \left({\displaystyle \frac{{10}^{-3}nm}{1pm}}\right)\right)}^2}{\left(1240eV.nm\right)\times \left({\left(4\right)}^2-{\left(1\right)}^2\right)} \\ =\frac{2,55,500eV.n{m}^2}{\left(1240ev.nm\right)\times \left(12\right)} \\ =17.2nm \end{gathered}$$

Therefore, the shortest wave length that can be is  $17.2nm$.

Step 6: (d) Find the longest wavelength:

For the longest wavelength the electron has moves down from  $n_{high}=2$ to$n_{low}=1$   quantum state.

 

Rewrite equation (2) as below.

 ${\lambda }_{high\to low}=\frac{8\left(m{c}^2\right)L^2}{hc\left({n^2}_{high}-{n^2}_{low}\right)}$

 

Substitute  $1240ev.nm for hc, 0.511MeV for m{c}^2,2 for n_{high}, 1for n_{low} and for 250pm for L$    in the above equation.

 

 $$\begin{gathered} {\lambda }_{2\to 1}=\frac{8\times 0.511MeV\times \left({\displaystyle \frac{{10}^{6}ev}{1MeV}}\right)\times {\left(250pm\times \left({\displaystyle \frac{{10}^{-3}nm}{1pm}}\right)\right)}^2}{\left(1240eV.nm\right)\times \left({\left(2\right)}^2-{\left(1\right)}^2\right)} \\ =\frac{2,55,500eV.n{m}^2}{\left(1240ev.nm\right)\times 3} \\ =68.7nm \end{gathered}$$

Therefore, the shortest wave length that can be is  $68.7nm$.

Step 7: (e) Find the second longest wavelengths:

The second longest wave length emitted when electron has moves down from  $n_{high}=3$ to  $n_{low}=2$ quantum state.

 

Rewrite equation (2) as follow.

 ${\lambda }_{high\to low}=\frac{8\left(m{c}^2\right)L^2}{hc\left({n^2}_{high}-{n^2}_{low}\right)}$

 

Substitute  $1240ev.nm for hc, 0.511MeV for m{c}^2, 3 for n_{high}, 2for n_{low} and for 250pm for L$   in the above equation.

 

 

 $$\begin{gathered} {\lambda }_{3\to 2}=\frac{8\times 0.511MeV\times \left({\displaystyle \frac{{10}^{6}ev}{1MeV}}\right)\times {\left(250pm\times \left({\displaystyle \frac{{10}^{-3}nm}{1pm}}\right)\right)}^2}{\left(1240eV.nm\right)\times \left({\left(3\right)}^2-{\left(2\right)}^2\right)} \\ =\frac{2,55,500eV.nm}{\left(1240ev.nm\right)\times 5} \\ =41.2nm \end{gathered}$$

Therefore, the shortest wave length that can be is $41.2nm$ .

Step 8: (f) Define the various possible ways on an energy-level :

The following figure shows the possible downwards transition of electron from third excited state to the ground state.

                                         

Step 9: (g) Find the longest wavelength:

Rewrite equation (2) and rearrange it in terms of  $n_{low}$.

 $$\begin{gathered} \left({n^2}_{high}-{n^2}_{low}\right)=\frac{8\left(m{c}^2\right)L^2}{hc\lambda } \\ {n^2}_{low}={n^2}_{high}=\frac{8\left(m{c}^2\right)L^2}{hc\lambda } \\ {n^2}_{low}=\left[{n^2}_{high}-\frac{8\left(m{c}^2\right)L^2}{hc\lambda }\right] \end{gathered}$$

 

Substitute  $1240ev.nm for hc, 0.511MeV for m{c}^2, 4 for n_{high}, 29.4nm for \lambda and for 250pm for L$  to find $lo{w}_{}$ .

 

 $$\begin{gathered} n_{low}={\left[{\left(4\right)}^2-\frac{8\times 0.511MeV\times \left({\displaystyle \frac{{10}^{6}eV}{1MeV}}\right)\times {\left(250pm\times \left({\displaystyle \frac{{10}^{-3}nm}{1pm}}\right)\right)}^2}{\left(1240eV.nm\right)\times 29.4nm}\right]}^{\frac{1}{2}} \\ ={\left[16-\frac{2,55,500eV.n{m}^2}{\left(1240eV.nm\right)\times 29.4nm}\right]}^{\frac{1}{2}} \\ ={\left[16-7\right]}^{\frac{1}{2}} \\ =3 \end{gathered}$$

 

Therefore, the emitted wave length 29.4 nm corresponds to transition from third excited state to second excited state  $4\to 3$.

Step 10: (h) Find the shortest wave- length that can be emitted:

The second longest wave length emitted when electron has moves down from $n_{high}=3$  to   $n_{low}=1$quantum state

Rewrite equation (2) as below.

${\lambda }_{high\to low}=\frac{8\left(m{c}^2\right)L^2}{hc\left({n^2}_{high}-{n^2}_{low}\right)}$ 

 

Substitute  $1240ev.nm for hc, 0.511MeV for m{c}^2, 3 for n_{high},1for n_{low} and for 250pm for L$    in the above equation.

 

 $$\begin{gathered} {\lambda }_{3\to 1}=\frac{8\times 0.511MeV\times \left({\displaystyle \frac{{10}^{6}ev}{1MeV}}\right)\times {\left(250pm\times \left({\displaystyle \frac{{10}^{-3}nm}{1pm}}\right)\right)}^2}{\left(1240eV.nm\right)\times \left({\left(3\right)}^2-{\left(1\right)}^2\right)} \\ =\frac{2,55,500eV.n{m}^2}{\left(1240eV.nm\right)\times 8} \\ =25.8nm \end{gathered}$$

Therefore, the shortest wave length that can be is  $25.8nm$.