91影视

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 =$1.6脳{10}^{-19}$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,

$p_f=p_{Fe.f}+p_{纬}$

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

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

$p_i=p_f=0$

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

The Energy for gamma ray is given by,

$E_{纬}=\frac{h}{位}c$

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

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

$P_{纬}=\frac{h}{位}$

And,

$E_{纬}=P_{纬}c$

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

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

$鈭�E=E_{纬}+K$

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

Substitute values in the above,

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

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

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

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

Substitute values in the above,

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

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

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

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

Substitute values in the above,

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

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 鈥淢ossbauer effect鈥� 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,

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

Thus, the recoil speed of the nucleus is $7.67脳{10}^{-24}\mathrm{m}/\mathrm{s}$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.