Question

If the potential energy of an electron in an orbit of hydrogen atom is $-6.80$ eV, then the energy required to ionize the atom is

• A. 6.80 eV
• B. 13.6 eV
• C. 3.40 eV
• D. 10.2 eV

Hint:

Remember: $TE = -13.6/n^2$. $PE = -27.2/n^2$. $KE = 13.6/n^2$. Here $PE = -6.8 \Rightarrow n=2$.

Answer 1

The Correct Option is C

Step 1: Relationship between Energies:
For an electron in a hydrogen atom: Potential Energy ($PE$) = $2 \times$ Total Energy ($TE$). Kinetic Energy ($KE$) = $-TE$. Relation: $PE = -2 KE = 2 TE$.
Step 2: Calculate Total Energy:
Given $PE = -6.80$ eV. \[ TE = \frac{PE}{2} = \frac{-6.80}{2} = -3.40 \, \text{eV} \]
Step 3: Calculate Ionization Energy:
Ionization Energy is the energy required to remove the electron from its orbit to infinity (where energy is 0). \[ \text{Ionization Energy} = 0 - (TE) \] \[ \text{Ionization Energy} = -(-3.40 \, \text{eV}) = 3.40 \, \text{eV} \]
Step 4: Final Answer:
The ionization energy is 3.40 eV.