Question

What is the energy of an electron in stationary state corresponding to n = 2?

• A. $-1.45 \times 10^{-18}\ \text{J}$
• B. $-0.545 \times 10^{-18}\ \text{J}$
• C. $-3.45 \times 10^{-18}\ \text{J}$
• D. $-2.5 \times 10^{-18}\ \text{J}$

Hint:

If you already remember the standard energy values in electron-volts, the calculation is even faster: $E_2 = -3.4\ \text{eV}$. Multiplying by the conversion factor ($1.6 \times 10^{-19}\ \text{J/eV}$) gives: $-3.4 \times 1.6 \times 10^{-19} = -5.44 \times 10^{-19}\ \text{J} = -0.544 \times 10^{-18}\ \text{J}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the quantitative total energy ($E_n$) of an electron moving inside the second stationary Bohr orbit ($n=2$) of a hydrogen atom, expressed in Joules.

Step 2: Key Formula or Approach:
According to Bohr's atomic model, the energy level of an electron residing in the $n$-th orbit of a hydrogen atom is defined by the relation: $$E_n = -\frac{2.18 \times 10^{-18}}{n^2}\ \text{J}$$

Step 3: Detailed Explanation:
We need to calculate the energy for the second orbit, so we substitute $n = 2$ into Bohr's energy formula: $$E_2 = -\frac{2.18 \times 10^{-18}}{2^2}\ \text{J}$$ $$E_2 = -\frac{2.18 \times 10^{-18}}{4}\ \text{J}$$ Performing the simple division: $$E_2 = -0.545 \times 10^{-18}\ \text{J}$$

Step 4: Final Answer:
The energy associated with the $n=2$ stationary state is $-0.545 \times 10^{-18}\ \text{J}$, which perfectly matches option (B).