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The total energy of an electron in the first excited state of hydrogen is about - 3.4 eV. Its kinetic energy in this state is :

  1. A
    - 3.4 eV
  2. B
    - 1.7 eV
  3. C
    1.7 eV
  4. D
    3.4 eV

Solution & Step-by-step Explanation

The total energy (E) of an electron in a hydrogen atom is the sum of its kinetic energy (KE) and potential energy (PE), given by E = KE + PE. For an electron in the first excited state (n = 2), the total energy is -3.4 eV. The potential energy of the electron in the nth state of a hydrogen atom is given by PE = -13.6 / n^2 eV, and for n = 2, PE = -13.6 / 4 = -3.4 eV. However, this is the total energy, not just the potential energy. The kinetic energy can be found from the relationship that the total energy is the negative of the kinetic energy for a bound system (since PE = -2KE for a bound electron in a hydrogen-like atom), thus KE = -E/2. For the first excited state, KE = -(-3.4 eV)/2 = 1.7 eV.

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The total energy of an electron in the first excited state of hydrogen is about - 3.4 eV. Its kinetic energy in this state is :
A
- 3.4 eV
B
- 1.7 eV
C
1.7 eV
D
3.4 eV

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