Question

An energy of 68 eV is required to excite a hydrogen-like atom from its second Bohr orbit to the third. The nuclear charge is Ze. The value of Z is:

• A. 6
• B. 4
• C. 3
• D. 2

Answer 1

Answer: (A)

6

$E=\frac{hc}{\lambda }=R{Z}^{2}hc[\frac{1}{n_1^2}-\frac{1}{n_1^2}]$
Electron transition is from $2\to 3$
$E=68eV=68\times 1.6\times {10}^{-19}J=R{Z}^{2}hc[\frac{1}{2^2}-\frac{1}{3^2}]$
$68\times 1.6\times {10}^{-19}J=\frac{5R{Z}^{2}hc}{36}$
$\Rightarrow Z^2=36$
$\Rightarrow Z=6$