Question

An electron revolves around a nucleus of charge $+Ze$ ( Z = 5), the energy required to excite the electron from $n=3$ to $n=4$ state will be

• A. 16.53 eV
• B. 13.6 eV
• C. 1.51 eV
• D. 27.2 eV

Answer 1

The correct option is A 16.53 eV

Energy needed to ionise an atom from $n^{th}$ orbit with charge Z = $13.6\times \frac{z^2}{n^2}$

Given:

$n_1=3$ to $n_2=4$

$z=5$

As we know ,

$E=13.6\times z^2\times (\frac{1}{n_1^2}-\frac{1}{n_2^2})$

putting the values , in above equation , we get

$E=13.6\times 5^2\times (\frac{1}{3^2}-\frac{1}{4^2})$

$E=13.6\times 25\times \frac{7}{9\times 16}$

After calculationg , we get

$E=16.53eV$

Hence , option A is correct .