Question

The ratio of $({\mathrm{E}}_2-{\mathrm{E}}_1)$ to $({\mathrm{E}}_4-{\mathrm{E}}_3)$ for the hydrogen atom is approximately equal to:

• A. $10$
• B. $15$
• C. $17$
• D. $12$

Answer 1

The correct option is B

$15$

Explanation for correct option:

(B) $15$

Step 1: Calculate the energy associated with the orbit of hydrogen:

$$\begin{gathered} {\mathrm{E}}_1=-\frac{13.6}{1^2}=-13.6[\mathrm{First}\mathrm{Orbit}] \\ {\mathrm{E}}_2=-\frac{13.6}{2^2}=-3.4[\mathrm{Second}\mathrm{Orbit}] \\ {\mathrm{E}}_3=-\frac{13.6}{3^2}=-1.51[\mathrm{Third}\mathrm{Orbit}] \\ {\mathrm{E}}_4=-\frac{13.6}{4^2}=-0.85[\mathrm{Fourth}\mathrm{Orbit}] \end{gathered}$$

Step 2: Calculation of the ratio:

The ratio $({\mathrm{E}}_2-{\mathrm{E}}_1)$ to $({\mathrm{E}}_4-{\mathrm{E}}_3)$ for hydrogen atom is approximately

$$\begin{gathered} \frac{{\mathrm{E}}_2-{\mathrm{E}}_1}{{\mathrm{E}}_4-{\mathrm{E}}_3}=\frac{-3.4-(-13.6)}{-0.85-(-1.51)} \\ \Rightarrow \frac{{\mathrm{E}}_2-{\mathrm{E}}_1}{{\mathrm{E}}_4-{\mathrm{E}}_3}=\frac{10.2}{0.66} \\ \Rightarrow \frac{{\mathrm{E}}_2-{\mathrm{E}}_1}{{\mathrm{E}}_4-{\mathrm{E}}_3}=15.4 \\ \Rightarrow \frac{{\mathrm{E}}_2-{\mathrm{E}}_1}{{\mathrm{E}}_4-{\mathrm{E}}_3}\approx 15 \end{gathered}$$

Therefore, the ratio of $({\mathrm{E}}_2-{\mathrm{E}}_1)$ to $({\mathrm{E}}_4-{\mathrm{E}}_3)$ for the hydrogen atom is approximately equal to $15$.

Explanation for incorrect options:

The ratio of $({\mathrm{E}}_2-{\mathrm{E}}_1)$ to $({\mathrm{E}}_4-{\mathrm{E}}_3)$ for the hydrogen atom is approximately equal to $15$. Thus, options A, C, and D are incorrect.

Hence, the correct option is B i.e. $15$.