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: Use de-Broglie wavelength relation.
\[ \lambda=\frac{h}{mv} \] Therefore, \[ mv=\frac{h}{\lambda} \]

Step 2: Write kinetic energy formula.
\[ K=\frac{1}{2}mv^2 \] Using \[ mv=\frac{h}{\lambda}, \] we get \[ K=\frac{1}{2m}\left(\frac{h}{\lambda}\right)^2 \] Hence, \[ K=\frac{h^2}{2m\lambda^2} \]

Step 3: Substitute the given values.
\[ h=6.6\times10^{-34}\,\text{Js} \] \[ m=9\times10^{-31}\,\text{kg} \] \[ \lambda=3.3\times10^{-10}\,\text{m} \] Thus, \[ K= \frac{(6.6\times10^{-34})^2} {2(9\times10^{-31})(3.3\times10^{-10})^2} \]

Step 4: Simplify the expression.
\[ (6.6)^2=43.56 \] \[ (3.3)^2=10.89 \] \[ K= \frac{43.56\times10^{-68}} {19.602\times10^{-50}} \] \[ K\approx2.22\times10^{-18}\,\text{J} \]

Step 5: Final conclusion.
Hence, the kinetic energy of the electron is \[ \boxed{2.22\times10^{-18}\,\text{J}} \]