Question

Question: You will find in Chapter 39 that electrons cannot move in definite orbits within atoms, like the planets in our solar system. To see why, let us try to “observe” such an orbiting electron by using a light microscope to measure the electron’s presumed orbital position with a precision of, say, $\mathbf{10}\mathbf{}\mathbf{pm}$(a typical atom has a radius of about localid="1663132292844" $\mathbf{100}\mathbf{}\mathbf{pm}$). The wavelength of the light used in the microscope must then be about $\mathbf{10}\mathbf{}\mathbf{pm}$. (a) What would be the photon energy of this light? (b) How much energy would such a photon impart to an electron in a head-on collision? (c) What do these results tell you about the possibility of “viewing” an atomic electron at two or more points along its presumed orbital path? (Hint:The outer electrons of atomsare bound to the atom by energies of only a few electron-volts.)

Short answer

(a) The photon energy of the light is$124\mathrm{KeV}$.

(b) The energy imparted by the photon to an electron in a head-on collision is$40.5\mathrm{KeV}$.

(c) According to the above results, we can conclude that making it is impossible to find an atomic electron with such high energy.

Step-by-step solution

Identifying the data given in the question

The wavelength of light $\lambda =10\mathrm{pm}$.

Collision of electron and photon is head therefore$\varphi ={180}^0$

Concept used to solve the question.

Photon Energy

The energy that a single photon carries is known as photon energy. Energy is inversely correlated with wavelength because it is directly proportional to the electromagnetic frequency of the photon.

(a) Finding the photon energy of light

Photon energy of light can be given as

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

Where $h$is plank constant, $v$is the velocity of light, and $\lambda$is the wavelength

We know $hc=1240nm·eV$

Substituting the values

$\begin{array}{rcl}E& =& \frac{1240nm·eV}{10\times {10}^{-3}nm}\\ & =& 124KeV\end{array}$

Hence the photon energy of the light is $124KeV$.

(b) Finding the energy imparted by the photon

As we know the energy gained by the Electron must be equal to the decrease in energy of the photon

So, the decrease in energy of the photon can be given as

$∆E=∆\left(\frac{hc}{\lambda }\right)$

Since $h$and $c$are constant

Therefore,

$\begin{array}{rcl}∆E& =& hc\left(\frac{1}{\lambda }-\frac{1}{\lambda +∆\lambda }\right)\\ & =& \frac{hc}{\lambda }\left(\frac{∆\lambda }{\lambda +∆\lambda }\right)\\ & =& \frac{E}{1+{\displaystyle \raisebox{1ex}{$\lambda $}\!\left/ \!\raisebox{-1ex}{$∆\lambda $}\right.}}\\ & =& \frac{E}{1+{\displaystyle \raisebox{1ex}{$\lambda $}\!\left/ \!\raisebox{-1ex}{${\lambda }_c\left(1-\cos \varphi \right)$}\right.}}\end{array}$

We know in the head-on collision$\varphi ={180}^{\circ }$

Now substituting the values

$\begin{array}{rcl}∆\mathrm{E}& =& \frac{\mathrm{E}}{1+{\displaystyle \raisebox{1ex}{$\mathrm{\lambda }$}\!\left/ \!\raisebox{-1ex}{${\mathrm{\lambda }}_{\mathrm{c}}$}\right.}\left(1-\mathrm{cos\varphi }\right)}\\ & =& \frac{124\mathrm{KeV}}{1+{\displaystyle \raisebox{1ex}{$10\mathrm{pm}$}\!\left/ \!\raisebox{-1ex}{$243\mathrm{pm}$}\right.}\left(1-\cos {180}^{\circ }\right)}\\ & =& 40.5\mathrm{KeV}\end{array}$

Hence the energy imparted by a photon to an electron in a head-on collision is $40.5\mathrm{KeV}$

 Step 5: (c) Conclusion from the above results

The energy imparted by a photon to an electron in a head-on collision is $40.5\mathrm{KeV}$

This is very high energy.

According to the above results, we can conclude that

With the energy supplied to the electron by the photon, the electron would have been thrown out of its orbit, making it impossible to find an atomic electron with such high energy.