Question

If the wavelenth of $K_{\alpha }$ radiation emitted by an atom of atomic number $Z=41$ is $\lambda$ , then the atomic number for an atom that emits $K_{\alpha }$ radiation with wavelength $4$ $\lambda$ , is

• A. $21$
• B. $28$
• C. $11$
• D. $36$

Answer 1

Answer: (A)

$21$

Answer 2

$\frac{1}{\lambda} \propto(Z-1)^{2}$ $\therefore \frac{\lambda_{1}}{\lambda_{2}}=\frac{\left(Z_{2}-1\right)^{2}}{\left(Z_{1}-1\right)^{2}}$ $\frac{1}{4}=\left(\frac{Z_{2}-1}{41-1}\right)^{2}$ $\frac{1}{2}=\frac{Z_{2}-1}{40}$ Solving this, we get, $Z_{2}=21$