91影视

Use the value formula then it gives by the de Broglie wavelength of the electron is:

$\begin{array}{c}位=\frac{h}{p}\\ =\frac{h}{\sqrt{2m_{e}K}}\\ =\frac{hc}{\sqrt{2m_{e}eV}}\\ =\frac{6.626脳{10}^{鈭�34}Js}{\sqrt{2(9.109脳{10}^{鈭�31}kg)(1.602脳{10}^{鈭�19}C)(2\stackrel{藱}{5}.0脳{10}^{3}V)}}\\ =7.75脳{10}^{鈭�12}m\\ =7.75\text{鈥夆赌媝尘}\end{array}$

Hence, the wavelength of thermal radiation is maximum at $7.75\text{鈥夆赌媝尘}$.

A photon鈥檚 de Broglie wavelength is equal to its familiar wave relationship value. Use the value$hc=1240\text{鈥媏痴}鈰�\text{nm}$ gives:

$\begin{array}{c}位=\frac{hc}{E}\\ =\frac{1240eV鈰�\text{鈥�}nm}{1.00\text{鈥�}keV}\\ =1.24\text{鈥夆赌媙尘}\end{array}$

Hence, the wavelength is $1.24\text{鈥夆赌媙尘}$.

The neutron mass equals to $1.675脳{10}^{鈭�27}kg$. Using the conversion from electronvolts to Joules gives;

$\begin{array}{c}位=\frac{h}{p}\\ =\frac{h}{\sqrt{2m_{e}K}}\\ =\frac{hc}{\sqrt{2m_{e}eV}}\\ =\frac{6.626脳{10}^{鈭�34}Js}{\sqrt{2(9.109脳{10}^{鈭�31}kg)(1.602脳{10}^{鈭�19}C)(2\stackrel{藱}{5}.0脳{10}^{3}V)}}\\ =7.75脳{10}^{鈭�12}m\\ =7.75\text{鈥夆赌媝尘}\end{array}$

Hence, the wavelength of thermal radiation is$7.75\text{鈥夆赌媝尘}$ .

A photon鈥檚 de Broglie wavelength is equal to its familiar wave relationship value. Use the value$hc=1240\text{鈥媏痴}鈰�\text{nm}$ gives:

$\begin{array}{c}位=\frac{hc}{E}\\ =\frac{1240eV鈰�\text{鈥�}nm}{1.00\text{鈥�}keV}\\ =1.24\text{鈥夆赌媙尘}\end{array}$

Hence, the wavelength is $1.24\text{鈥夆赌媙尘}$.