Q2 / 160 · TS EAMCET 2023 (14 May Shift 2) · Chemistry
The ratio of the difference in energy between the first and second Bohr orbits to that between the second and third orbits is
- A$\frac{5}{27}$
- Correct answerB$\frac{27}{5}$
- C$\frac{4}{9}$
- D$\frac{9}{4}$
Solution
The difference in energy between the first and second Bohr orbit is
$\begin{aligned} & E_2-E_1=-13.6 \times z^2 \times\left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right) \\ & =-13.6 \times(1)^2 \times\left(\frac{1}{2^2}-\frac{1}{1^2}\right)\end{aligned}$
$=-13.6 \times\left(-\frac{3}{4}\right)=13.6 \times \frac{3}{4}$ ...(i)
The difference in energy between the second and third Bohr orbit is
$E_3-E_2=-13.6 \times z^2 \times\left(\frac{1}{n_3^2}-\frac{1}{n_2^2}\right)$
$=-13.6 \times(1)^2 \times\left(\frac{1}{3^2}-\frac{1}{2^2}\right)$
$=-13.6 \times\left(\frac{-5}{36}\right)=13.6 \times \frac{5}{36}$ ...(ii)
$\therefore$ Ratio $=\frac{E_2-E_1}{E_3-E_2}=\frac{\times \frac{3}{4}}{\times \frac{5}{36}}=\frac{3}{4} \times \frac{36}{5}=\frac{27}{5}$
Hence, the correct answer is option: 'B'.