![If lambda(Cu) is the wavelength of K(alpha) X-ray line of copper (atomic number 29) and lambda(Mo) is the wavelength of the K(alpha)X-ray line of molybdenum (atomic number 42),then the ratio lambda(Cu)//lambda(Mo) is If lambda(Cu) is the wavelength of K(alpha) X-ray line of copper (atomic number 29) and lambda(Mo) is the wavelength of the K(alpha)X-ray line of molybdenum (atomic number 42),then the ratio lambda(Cu)//lambda(Mo) is](https://d10lpgp6xz60nq.cloudfront.net/ss/web/400750.jpg)
If lambda(Cu) is the wavelength of K(alpha) X-ray line of copper (atomic number 29) and lambda(Mo) is the wavelength of the K(alpha)X-ray line of molybdenum (atomic number 42),then the ratio lambda(Cu)//lambda(Mo) is
![K(alpha) radiation of Mo (Z = 42) has a wavelength of 0.71Å. Calculate wavelength of the corresponding radiation of Cu (Z=29) K(alpha) radiation of Mo (Z = 42) has a wavelength of 0.71Å. Calculate wavelength of the corresponding radiation of Cu (Z=29)](https://d10lpgp6xz60nq.cloudfront.net/web-thumb/415585498_web.png)
K(alpha) radiation of Mo (Z = 42) has a wavelength of 0.71Å. Calculate wavelength of the corresponding radiation of Cu (Z=29)
![electromagnetism - Why is $K_{\alpha,3/2}$ always more intense than $K_{\ alpha,1/2}$ in copper? - Physics Stack Exchange electromagnetism - Why is $K_{\alpha,3/2}$ always more intense than $K_{\ alpha,1/2}$ in copper? - Physics Stack Exchange](https://i.stack.imgur.com/GH8Ri.png)
electromagnetism - Why is $K_{\alpha,3/2}$ always more intense than $K_{\ alpha,1/2}$ in copper? - Physics Stack Exchange
Calculate the wavelength of kalpha line for Z = 31 when alpha = 5 × 10^7Hz^1/2 for a characteristic X - ray spectrum.
![If `lambda_(Cu)` is the wavelength of `K_(alpha)` X-ray line of copper (atomic number `29`) - YouTube If `lambda_(Cu)` is the wavelength of `K_(alpha)` X-ray line of copper (atomic number `29`) - YouTube](https://i.ytimg.com/vi/Eo5lYTogdN0/maxresdefault.jpg)