Ncert Exercises & Solutions

12.15 The total energy of an electron in the first excited state of the hydrogen atom is about –3.4 eV.

(a) What is the kinetic energy of the electron in this state?

(b) What is the potential energy of the electron in this state?

(a)
Hint: Kinetic energy= -Total energy.
Step 1: Find the kinetic energy of the electron.

Total energy of the electron, E= -3.4 eV
The kinetic energy of the electron is equal to the negative of the total energy.
K= - E
-(-3.4)= +3.4 eV
Hence, the kinetic energy of the election in the given state is +3.4 eV.
(b)
Hint: The magnitude of potential energy of an electron is double of the amgnitude of its kinetic energy.
Step 2: Find the potential energy of the electron.
Potential energy (U) of the electron is equal to the negative of twice of its kinetic energy.
U=-2K = -2x3.4 = -6.8 eV
Hence, the potential energy of the election in the given state is -6.8 eV.
(c)
Hint: Potential energy depends on the reference point.
Step 3: Find the change in potential energy if the reference frame is changed.

The potential energy of a system depends on the reference point taken. Here, the potential energy of the reference point is taken as zero. If the reference point is changed, then the value of the potential energy of the system also changes.
Since total energy is the sum of kinetic and potential energies, the total energy of the system will also change.

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