91影视

(a)

The mass of an electron is $m=9.109脳{10}^{鈭�31}\text{kg}$, its Compton wavelength is solved as follows:

$\begin{array}{c}{位}_C=\frac{h}{mc}\\ =\frac{6.626脳{10}^{鈭�34}\text{J.s}}{(9.109脳{10}^{鈭�31}\text{kg})(2.998脳{10}^8\text{m/s})}\\ =2.426脳{10}^{鈭�12}\text{m}\\ =2.43\text{鈥塸尘}\end{array}$

Hence, the Compton wavelength for an electron is$2.43\text{鈥塸尘}$ .

(b)

The mass of a proton is$m=1.673脳{10}^{鈭�27}\text{kg}$, its Compton wavelength is solved as follows:

$\begin{array}{c}{位}_C=\frac{h}{mc}\\ =\frac{6.626脳{10}^{鈭�34}\text{J.s}}{(1.673脳{10}^{鈭�27}\text{kg})(2.998脳{10}^8\text{m/s})}\\ =1.321脳{10}^{鈭�15}\text{m}\\ =1.32\text{鈥塮尘}\end{array}$

Hence, the Compton wavelength for a proton is $1.32\text{鈥塮尘}$.

(c)

Let $hc=1240\text{鈥塭痴}鈰�\text{nm}$then it gives,

$\begin{array}{c}E=\frac{hc}{位}\\ =\frac{1240\text{鈥塶尘}鈰�\text{eV}}{位}\end{array}$

Here, E is the energy and $位$is the wavelength.

Thus, the energy for the electron is;

$\begin{array}{c}E=\frac{1240\text{鈥塶尘}鈰�\text{eV}}{2.426脳{10}^{鈭�3}\text{nm}}\\ =5.11脳{10}^5\text{鈥�}eV\\ =0.511\text{鈥塎别痴}\end{array}$

Hence, the energy for the electron is$0.511\text{鈥塎别痴}$ .

(d)

Let $hc=1240\text{鈥塭痴}鈰�\text{nm}$then it gives,

$\begin{array}{c}E=\frac{hc}{位}\\ =\frac{1240\text{鈥塶尘}鈰�\text{eV}}{位}\end{array}$

Here, E is the energy and $位$is the wavelength.

Thus, the energy for the proton is;

$\begin{array}{c}E=\frac{1240\text{鈥塶尘}鈰�\text{eV}}{1.321脳{10}^{鈭�6}\text{鈥塶尘}}\\ =9.39脳{10}^8\text{鈥�}eV\\ =939\text{鈥塎别痴}\end{array}$

Hence, the energy for the proton is$939\text{鈥塎别痴}$ .