Question

What is the energy of a hydrogen atom in the first excited state if the potential energy is taken to be zero on the ground state?

Answer 1

Step 1: Using Bohr's theory to calculate energy

• The spectrum of hydrogen atoms is explained by the Bohr hypothesis of atomic structure.
• It is predicted that an electron orbiting a nucleus may only exist in a few different energy levels, with each transition accompanied by a quantum of radiation being emitted or absorbed.
• To break a hydrogen atom apart or to split the proton and electron such they are infinitely far apart from one another, 13.6 eV of energy must be supplied.
• Energy required to move atom from n₁ state to n₂ state, $\mathrm{E}=13.6\times \left(\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right)\mathrm{eV}$
• Here, n₁= 1 (ground state),
• n₂= 2 (first excited state)

Step 2 : putting values in the formula

• $\mathrm{E}=13.6\times \left(\frac{1}{1_{}^2}-\frac{1}{2_{}^2}\right)\mathrm{eV}$
• $\mathrm{E}=13.6\times \left(\frac{1}{1_{}^{}}-\frac{1}{4_{}^{}}\right)\mathrm{eV}$
• $\mathrm{E}=13.6\times \left(\frac{3}{4}\right)\mathrm{eV}$
• $\mathrm{E}=10.2\mathrm{eV}$

Step 3:Total energy in the first excited state:

• $13.6eV+10.2eV=23.8eV$