A Fe-57 nucleus is at rest and in its first excited state, 14.4 keV above the ground state (14.4 × 103 eV, where 1 eV = 1.6×10−19 J). The nucleus then decays to the ground state with the emission of a gamma ray (a high-energy photon). (a) Wthe recoil speed of the nucleus? (b) Calculate the slight difference in eV between the gamma-ray energy and the 14.4 keV difference between the initial and final nuclear states. (c) The “Mössbauer effect” is the name given to a related phenomenon discovered by Rudolf Mössbauer in 1957, for which he received the 1961 Nobel Prize for physics. If the Fe-57 nucleus is in a solid block of iron, occasionally when the nucleus emits a gamma ray the entire solid recoils as one object. This can happen due to the fact that neighbouring atoms and nuclei are connected by the electric interatomic force. In this case, repeat the calculation of part (a) and compare with your previous result. Explain briefly

The given data can be listed below,

The energy of the first excited state of Fe-57 nucleus is, .

The value of one electron volt is,1eV =$1.6\times {10}^{-19}$J.

The momentum of any object is mathematically described as the product of mass of the object and its velocity.

The initial momentum of the Fe nucleus is zero.

$p_i=0$

The final momentum is given by,

$p_f=p_{Fe.f}+p_{\gamma }$

Here, $p_{Fe.f}$and $p_{\gamma }$are the final momentum of the Fe nucleus and the Gamma ray respectively.

The system is conserved so the net force is zero. So,

$p_i=p_f=0$

$\left|P_{Fe,f}\right|=\left|P_{\gamma }\right|$ ...(i)

The Energy for gamma ray is given by,

$E_{\gamma }=\frac{h}{\lambda }c$

Here, his the Planck’s constant whose value is $6.626\times {10}^{-34}\mathrm{J}-\mathrm{s}$,$\lambda$ is the De-Broglie wavelength and cis the speed of light

From De-Broglie Wavelength, momentum of the gamma ray is given by,

$P_{\gamma }=\frac{h}{\lambda }$

And,

$E_{\gamma }=P_{\gamma }c$

Again, from equation (1),$E_{\gamma }=P_{Fe,f}c$

Recoil Kinetic Energy,

$K=\frac{1}{2}m{v}_{Fe,f}^2$ ...(ii)

Where,$v_{Fe,f}$is the recoil speed of the nucleus.

role="math" localid="1657867555746" $v_{Fe,f}=\frac{P_{Fe,f}}{m}$ ...(iii)

Substitute the above value in (ii)

$K=\frac{P_{Fe,f^2}}{2m}$

The energy difference expression is

$∆E=E_{\gamma }+K$

Here, $E_{\gamma }$is the energy of gamma ray and Kis the recoil kinetic energy of the iron nucleus.

Substitute values in the above,

$$\begin{gathered} ∆\mathrm{E}=14.4\times {10}^3\mathrm{eV}\left(\frac{1.6\times {10}^{-19}\mathrm{J}}{1\mathrm{eV}}\right) \\ =2.3\times {10}^{-15}\mathrm{J} \end{gathered}$$

The value of the final momentum of the iron nucleus is given by,

$∆E=P_{Fe,f}c+\frac{{P_{Fe,f}}^2}{2m}$

Here, mis the mass of the nucleus whose value is$\mathrm{m}=9.52\times {10}^{-26}\mathrm{kg}$ and c is the speed of the light whose value is$\mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s}$

Substitute values in the above,

$$\begin{gathered} 2.3\times {10}^{-15}J=P_{Fe,f}c+\frac{P_{Fe,f}^2}{2m} \\ 2m\left(2.3\times {10}^{-15}J\right)=P_{Fe,f}c\left(2m\right)+P_{Fe,f}^2 \\ {P}_{Fe,f}=7.67\times {10}^{-24}kg▯m/s \end{gathered}$$

Substitute all the values in equation (iii), the recoil speed of the nucleus is given by,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{Fe},\mathrm{f}}=\frac{7.67\times {10}^{-24}\mathrm{kg}.\mathrm{m}/\mathrm{s}}{9.25\times {10}^{-26}\mathrm{kg}} \\ =80.6\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the recoil speed of the nucleus is 80.6 m/s .

The kinetic energy expression is given by,

$K=\frac{P_{Fe,f}^2}{2m}$

Here, m is the mass of the nucleus whose value is $\mathrm{m}=9.52\times {10}^{-26}\mathrm{kg}$, and$P_{Fe,f}$ is the final momentum of the nucleus.

Substitute values in the above,

$$\begin{gathered} \mathrm{K}=\frac{{\left(7.67\times {10}^{-24}\mathrm{kg}.\mathrm{m}/\mathrm{s}\right)}^2}{9.52\times {10}^{-26}\mathrm{kg}} \\ =0.0019\mathrm{eV} \end{gathered}$$

Hence, the difference between gamma ray energy and first excited state is 0.0019 eV .

The recoil speed of the iron block is given by,

${v^{\text{'}}}_{Fe,f}=\frac{P_{Fe,f}}{m}$

Here, from “Mossbauer effect” mis the mass of iron block whose value is 1kg and$P_{Fe,f}$ is the momentum of iron.

Substitute all the values in the above,

$$\begin{gathered} {{\mathrm{v}}^{\text{'}}}_{\mathrm{Fe},\mathrm{f}}=\frac{7.67\times {10}^{-24}\mathrm{kg}\square \mathrm{m}/\mathrm{s}}{1\mathrm{kg}} \\ =7.67\times {10}^{-24}\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the recoil speed of the nucleus is $7.67\times {10}^{-24}\mathrm{m}/\mathrm{s}$and it is seen from above results that the recoil speed of the nucleus is greater in value than the recoil speed of entire iron block.