Question

In hydrogen atom, an electron is transferred from an orbit of radius $1.3225$ nm to another orbit of radius $0.2116$ nm. What is the energy (in J) of emitted radiation? (Rydberg constant $R_H \approx 1.097 \times 10^7 \text{ m}^{-1}$)

• A. $1.635\times 10^{-18}$
• B. $3.027\times 10^{-19}$
• C. $4.087\times 10^{-19}$
• D. $0.4578\times 10^{-18}$

Hint:

The radius of the $n^{th}$ orbit in a hydrogen atom is $r_n = n^2 a_0$. For an electron transition, the energy of the emitted/absorbed photon is $E = 13.6 \left|\frac{1}{n_f^2} - \frac{1}{n_i^2}\right| \text{ eV}$. To convert to Joules, multiply by $1.602 \times 10^{-19} \text{ J/eV}$.

Answer 1

The Correct Option is D

Step 1: Relate the radius of the orbit to the principal quantum number $n$.
The radius of the $n^{th}$ orbit in a hydrogen atom is given by $r_n = n^2 a_0$, where $a_0$ is the Bohr radius, $a_0 \approx 0.0529 \text{ nm}$. 
Let $r_1 = 1.3225 \text{ nm}$ and $r_2 = 0.2116 \text{ nm}$. 
For the initial orbit ($n_1$): $n_1^2 = \frac{r_1}{a_0} = \frac{1.3225 \text{ nm}}{0.0529 \text{ nm}} = 25 \implies n_1 = 5$. 
For the final orbit ($n_2$): $n_2^2 = \frac{r_2}{a_0} = \frac{0.2116 \text{ nm}}{0.0529 \text{ nm}} = 4 \implies n_2 = 2$. 
The transition is from $n_1=5$ to $n_2=2$. Since $n_1>n_2$, a photon is emitted. 

Step 2: Calculate the energy of the emitted photon in $\text{eV$.} 
The energy of the emitted photon, $E$, is the difference in the energy levels: \[ E = E_{n_1} - E_{n_2} = -13.6 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \text{ eV} = 13.6 \left(\frac{1}{n_2^2} - \frac{1}{n_1^2}\right) \text{ eV}. \] Substitute $n_1=5$ and $n_2=2$: \[ E = 13.6 \left(\frac{1}{2^2} - \frac{1}{5^2}\right) \text{ eV} = 13.6 \left(\frac{1}{4} - \frac{1}{25}\right) \text{ eV}. \] \[ E = 13.6 \left(\frac{25-4}{100}\right) \text{ eV} = 13.6 \times \frac{21}{100} = 0.136 \times 21 \text{ eV}. \] \[ E = 2.856 \text{ eV}. \] 

Step 3: Convert the energy from $\text{eV$ to Joules (J).} 
The conversion factor is $1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$. We will use $1.6 \times 10^{-19} \text{ J}$ for simplification, or the exact value to match the option. Let's use the given option's implied conversion factor. \[ E (\text{J}) = 2.856 \text{ eV} \times (1.6 \times 10^{-19} \text{ J/eV}). \] \[ E (\text{J}) = 4.5696 \times 10^{-19} \text{ J}. \] In the form $0.4578 \times 10^{-18} \text{ J}$, this is: \[ E (\text{J}) = 0.45696 \times 10^{-18} \text{ J}. \] This is approximately $0.4578 \times 10^{-18} \text{ J}$ (Option D).