Question

For the hydrogen atom, $E_n=-\frac{1}{n^2}{E}_1$, Where, $E_1=2.178\times {10}^{-18}J$.
Assuming that the electrons in other shells exert no effect, find the minimum no. of $Z$ (atomic no.) at which a transition from the second energy level to the first results in the emission of an X-ray.
Given that the lowest energy X-rays have $\lambda =4\times {10}^{-8}m$.

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is A $2$

$E_1=2.178\times {10}^{-18}J$

$E_2=\frac{2.178\times {10}^{-18}}{4}{E}_2-E_1=\frac{hc}{\lambda }$

$Z^2\times 2.178\times {10}^{-18}\left(\frac{3}{4}\right)=\frac{3\times {10}^8\times 6.626\times {10}^{-34}}{4\times {10}^{-8}}$

$Z^2=4$

Hence $Z=2$