Question

An electron in a hydrogen-like atom is in an excited state. It has total energy of -3.4 eV. The kinetic energy and de-Broglie wavelength respectively are:

• A. 3.4 eV, 0.66 nm
• B. -3.4 eV, 1.99 nm
• C. 2.8 eV, 2.38 nm
• D. 1.1 eV, 1.28 nm

Hint:

For hydrogen atom, kinetic energy = magnitude of total energy.

Answer 1

The Correct Option is A

Step 1: Relation of energies. E = -K K = 3.4 eV
Step 2: de-Broglie wavelength. λ = h√(2mK) ≈ 0.66 nm