Question

Estimate the difference in energy between $1^{st}$ and $2^{nd}$ Bohr orbit for a $H$-atom. At what minimum atomic number, a transition from $n=2$ to $n=1$ energy level would result in the emission of $X$-ray with $\lambda =3.0\times {10}^{-8}m$. Which hydrogen atom-like species does this atomic number correspond to?

• A. $Z=2$
• B. $Z=3$
• C. $Z=4$
• D. None of these

Answer 1

Answer: (A)

$Z=2$

$E_1$ for $H=-13.6eV$

$\therefore E_2$ for $H=-(-13.6/2^2)=-13.6/4=-3.4eV$

$\therefore E_2-E_1=-3.4-(-13.6)=+10.2eV$

Also for transition of $H$ like atom; $\lambda =3.0\times {10}^{-8}m$

$\frac{1}{\lambda }=R_H{Z}^2[\frac{1}{1^2}-\frac{1}{2^2}]$

$\frac{1}{3\times {10}^{-8}}=1.09\times {10}^7\times Z^2\times \frac{3}{4}$

$\therefore Z^2=4$ and $Z=2$

Thus, the $H$ like an atom is $H{e}^{+}$.

Hence, the correct option is $A$