Question

The wavelength of electron in the first orbit of hydrogen atom is \[ 3.3\times10^{-10}\,\text{m}. \] The kinetic energy of electron (in J) is:
\[ (h=6.6\times10^{-34}\,\text{Js},\ m_e=9.0\times10^{-31}\,\text{kg}) \]

• A. \(3.33\times10^{-17}\)
• B. \(1.11\times10^{-18}\)
• C. \(2.22\times10^{-18}\)
• D. \(2.22\times10^{-17}\)

Hint:

For matter wave problems, directly use: \[ K=\frac{h^2}{2m\lambda^2} \] to quickly calculate kinetic energy.

Answer 1

The Correct Option is C

Step 1: Recall the de Broglie relation and orbit condition.
For an electron in the first orbit of hydrogen, the de Broglie wavelength equals the circumference: $\lambda = 2\pi r_1$. Here we are given $\lambda = 3.3 \times 10^{-10}$ m directly.
Step 2: Write the momentum from de Broglie relation.
\[ p = \frac{h}{\lambda} = \frac{6.6 \times 10^{-34}}{3.3 \times 10^{-10}} = 2.0 \times 10^{-24} \text{ kg m s}^{-1} \]
Step 3: Calculate kinetic energy using momentum.
Kinetic energy is related to momentum by: \[ KE = \frac{p^2}{2m_e} \] This avoids needing velocity directly.
Step 4: Substitute values.
\[ KE = \frac{(2.0 \times 10^{-24})^2}{2 \times 9.0 \times 10^{-31}} \] \[ KE = \frac{4.0 \times 10^{-48}}{1.8 \times 10^{-30}} \]
Step 5: Perform the division.
\[ KE = \frac{4.0}{1.8} \times 10^{-48+30} = 2.22 \times 10^{-18} \text{ J} \]
Step 6: Verify the units and result.
$p^2$ has units kg$^2$ m$^2$ s$^{-2}$; dividing by kg gives kg m$^2$ s$^{-2}$ = J. The result is consistent with the known ground-state energy of hydrogen (~13.6 eV = 2.18 $\times 10^{-18}$ J).
\[ \boxed{KE = 2.22 \times 10^{-18} \text{ J}} \]