Question

The minimum atomic number for which a transition from n = 2 to n = 1 energy level would result in the emission of X-rays with $\lambda =3.0\times {10}^{-8}m$ is:

• A. Z = 4
• B. Z = 2
• C. Z = 1
• D. Z = 3

Answer 1

The correct option is B Z = 2

For a transition from $n_2$ = 2 to $n_1$ = 1, wavenumber can be written as:
$\frac{1}{\lambda }=R_H{Z}^2[\frac{1}{1^2}-\frac{1}{2^2}]=R_H{Z}^2\times \frac{3}{4}$
Given $\lambda =3.0\times {10}^{-8}m$
$\frac{1}{3.0\times {10}^{-8}}=1.0967\times {10}^7\times Z^2\times \frac{3}{4}$
$\therefore Z^2=\frac{{10}^8\times 4}{3\times 3\times 1.0967\times {10}^7}=\frac{40}{9\times 1.0967}=4$
$\therefore Z=2$