Question

$K_{\alpha }$ wavelength emitted by an atom of atomic number $Z=11$ is ${}^{\prime }\lambda {}^{\prime }$. The atomic number for an atom that emits $K_{\alpha }$ radiation with wavelength $4\lambda$ is

• A. $6$
• B. $4$
• C. $11$
• D. $44$

Answer 1

The correct option is A $6$

$K_{\alpha }$ corresponds to $n=2\to n=1$
$\frac{1}{\lambda }=Z^{2}R(1-\frac{1}{n^2})$
$\frac{1}{\lambda }=121R(1-\frac{1}{2^2})$
$\frac{1}{\lambda }=121R\left(\frac{3}{4}\right)$ ---------------(1)
$\frac{1}{4\lambda }=Z^{2}R\left(\frac{3}{4}\right)$ ----------(2)

Dividing (1) by (2)
$4Z^2=121$
$Z=5.5$