For hydrogen atom, energy of an electron in first excited state is -3.4 eV, K . E. of the same electron of hydrogen atom is x eV. Value of x is _________ × 10⁻¹ eV. (Nearest integer)

Solution

Chemistry ma’am · 3 steps

In a Bohr orbit, kinetic energy is the total energy with its sign flipped, so just drop the minus sign.

1. Energy relations. For an electron in a Bohr orbit, the potential energy is twice the total energy and the kinetic energy is minus the total energy. So KE = -E and PE = 2E.

   KE = -Eₜₒₜₐₗ, PE = 2Eₜₒₜₐₗ

2. Put in numbers. The first excited state is n = 2, with E = -3.4 eV. So KE = -(-3.4) = 3.4 eV.
3. So. x = 3.4 eV = 34 × 10⁻¹ eV, so the answer is 34.

The common slip: -34. Copying the total energy without flipping its sign gives -34, but kinetic energy can never be negative.

The idea to keep. For any Bohr orbit, remember KE : PE : E = 1 : -2 : -1. Kinetic energy is always positive and equals the magnitude of the total energy.

Answer from the NTA final answer key. Solution by PrepWiser.