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The total energy E of an electron in a hydrogen atom is the sum of its kinetic energy K and its potential energy U. According to the Bohr model of the hydrogen atom, the total energy of the electron is given by:
E = K + U
The kinetic energy K is equal in magnitude to the potential energy, but opposite in sign. Therefore, we have:
K = -U
The total energy of the electron is:
E = K + (-K) = -K
Thus, the ratio of the total energy E to the kinetic energy K is:
E / K = -K / K = -1
Correct Answer: (A) -1
Step 1: Consider the total energy and kinetic energy of an electron in a hydrogen atom using Bohr's model.
The electron experiences an electrostatic force due to the nucleus, which provides the centripetal force required for circular motion: \[ \frac{k e^2}{r^2} = \frac{mv^2}{r} \]
Step 2: Multiply both sides by \( r \): \[ \frac{k e^2}{r} = mv^2 \]
Kinetic energy of the electron is: \[ K = \frac{1}{2}mv^2 = \frac{1}{2} \cdot \frac{k e^2}{r} = \frac{k e^2}{2r} \]
Step 3: Potential energy of the electron in the electrostatic field is: \[ U = -\frac{k e^2}{r} \]
Total energy of the electron is: \[ E = K + U = \frac{k e^2}{2r} - \frac{k e^2}{r} = -\frac{k e^2}{2r} \]
Step 4: Now take the ratio of total energy to kinetic energy: \[ \frac{E}{K} = \frac{-\frac{k e^2}{2r}}{\frac{k e^2}{2r}} = -1 \]
Final Answer: -1
If the voltage across a bulb rated 220V – 60W drops by 1.5% of its rated value, the percentage drop in the rated value of the power is: