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NEET CHEMISTRYEasy

In the Hydrogen atom, the energy of the first excited state is – 3.4 eV. Then find out the kinetic energy (K.E.) of the same orbit of the Hydrogen atom:

A
  • 3.4 eV
B
  • 6.8 eV
C

– 13.6 eV

D
  • 13.6 eV

Step-by-Step Solution

In the Bohr model of the hydrogen atom, the electron revolves around the nucleus under the influence of electrostatic attraction. For a system bound by an inverse square law force (like electrostatic or gravitational force), the relationship between Total Energy (EE), Kinetic Energy (KK), and Potential Energy (UU) is given by the virial theorem: E=KE = -K U=2EU = 2E

Given the total energy of the first excited state is E=3.4 eVE = -3.4 \text{ eV}.

To find the Kinetic Energy (KK): K=EK = -E K=(3.4 eV)K = -(-3.4 \text{ eV}) K=+3.4 eVK = +3.4 \text{ eV}

Thus, the kinetic energy is equal in magnitude to the total energy but is always positive .

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