Question

Match List I with List II:

LIST I	LIST II
A. Energy of ground state of He	I. \( +6.04 \, \text{eV} \)
B. Potential energy of I orbit of H-atom	II. \( -27.2 \, \text{eV} \)
C. Kinetic Energy of II excited state of He	III. \( 54.4 \, \text{eV} \)
D. Ionization potential of He	IV. \( -54.4 \, \text{eV} \)

Choose the correct answer from the options given below:

• A. (A)- (IV), (B)- (II), (C)- (I), (D)- (III)
• B. (A)- (I), (B)- (III), (C)- (II), (D)- (IV)
• C. (A)- (I), (B)- (II), (C)- (IV), (D)- (III)
• D. (A)- (III), (B)- (IV), (C)- (I), (D)- (II)

Hint:

Match the correct energy values with their corresponding quantum states based on known data for elements.

Answer 1

The Correct Option is A

To correctly match the items from List I with List II, we need to understand the energy values associated with specific states and properties of hydrogen (H) and helium (He) atoms.

1. Understanding the Given Quantities:
   
   • Energy of ground state of He (A): This is the total energy of a helium atom in its lowest energy state.
   • Potential energy of 1s orbit of H-atom (B): This refers to the potential energy of the electron in the first energy level (1s) of a hydrogen atom.
   • Kinetic Energy of II excited state of He (C): This is the kinetic energy of an electron in the second excited state of a helium atom.
   • Ionization potential of He (D): The energy required to remove an electron from a helium atom.
2. Known Energy Values:
   
   • Hydrogen Atom:
     
     • Ground state energy: \( E_n = -13.6 \, \text{eV} \, \text{for} \, n = 1 \).
     • Potential energy (\( PE \)) is twice the total energy: \( PE = 2 \times (-13.6) = -27.2 \, \text{eV} \).
     • Kinetic energy (\( KE \)) is equal in magnitude but opposite in sign to potential energy: \( KE = +13.6 \, \text{eV} \).
   • Helium Atom:
     
     • Ground state energy: \( E = -54.4 \, \text{eV} \).
     • Ionization potential: \( 24.6 \, \text{eV} \) (energy required to remove one electron).
     • For kinetic energy of an excited state, specific values depend on the energy level, but given the options, it aligns with \( +6.04 \, \text{eV} \).
3. Matching the Items:
   
   • A. Energy of ground state of He:
     Ground state energy of He = \( -54.4 \, \text{eV} \) (IV)
   • B. Potential energy of 1s orbit of H-atom:
     \( PE = -27.2 \, \text{eV} \) (II)
   • C. Kinetic Energy of II excited state of He:
     \( KE = +6.04 \, \text{eV} \) (I)
   • D. Ionization potential of He:
     Ionization potential = \( 24.6 \, \text{eV} \) (III)

Therefore, the correct matching is:
(A) - (IV), (B) - (II), (C) - (I), (D) - (III).