Question:
The speed of an electron in an orbit of hydrogen atom is \( \frac{1}{274} \) times the speed of light in vacuum. The kinetic energy of the electron in the orbit is:
The speed of an electron in an orbit of hydrogen atom is \( \frac{1}{274} \) times the speed of light in vacuum. The kinetic energy of the electron in the orbit is:
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The kinetic energy of an electron in the first orbit of a hydrogen atom is \( 3.4 \, \text{eV} \), which is half of the total energy (\(-13.6 \, \text{eV}\)).
Updated On: May 14, 2025
- 3.4 eV
- 10.2 eV
- 13.6 eV
- 5.4 eV
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The Correct Option is A
Solution and Explanation
The kinetic energy of the electron in the first orbit of hydrogen is given by:
\[
K.E. = \frac{1}{2} mv^2
\]
From Bohr's model, the total energy of the electron in the first orbit is \( -13.6 \, \text{eV} \). The kinetic energy is half of the total energy, and its positive value is:
\[
K.E. = 3.4 \, \text{eV}
\]
Thus, the kinetic energy of the electron in the first orbit is \( 3.4 \, \text{eV} \).
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