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

The correct option is B 6

As per question,

given that atomic number of an element $Z_1=11$

a wavelength of an element${\lambda }_1=\lambda$

the atomic number of the second element $Z_2=?$

a wavelenght of the second element ${\lambda }_2=4\lambda$

According to Moseley's law $\sqrt{v}=a(z-b)$

or $v=a^2(z-b{)}^2$
or $\frac{c}{\lambda }=a^2(z-b{)}^2$
$\therefore \frac{{\lambda }_1}{{\lambda }_2}=\frac{(z_2-1{)}^2}{(z_1-1{)}^2}$
$\therefore \frac{\lambda }{4\lambda }=\frac{(z_2-1{)}^2}{(11-1{)}^2}$
or $(z_2-1{)}^2=25$ or $z_2=6$

So the atomic number of the second element$Z_2=6$