Question

An element with atomic number $Z = 11$ emits $K_{\alpha} -$ X-ray of wavelength $\lambda$. The atomic number which emits $K_{\alpha} -$ $X$-ray of wavelength 4$\lambda$ is

• A. 4
• B. 6
• C. 11
• D. 44

Answer 1

Answer: (B)

6

Answer 2

According to Moseley's law $\sqrt{v}=a(z-b)$
or $v=a^{2}(z-b)^{2}$
or $\frac{c}{\lambda}=a^{2}(z-b)^{2}$
$\therefore \frac{\lambda_{1}}{\lambda_{2}}=\frac{\left(z_{2}-1\right)^{2}}{\left(z_{1}-1\right)^{2}}$
Here $\lambda_{1}=\lambda, \lambda_{2}=4 \lambda, z_{1}=11$ and $z_{2}=?$
$\therefore \frac{1}{4 \lambda}=\frac{\left(z_{2}-1\right)^{2}}{(11-1)^{2}}$
or $\left(z_{2}-1\right)^{2}=25$ or $z_{2}=6$