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  • Stuck here, help me understand: - Atomic Structure - JEE Main-2

Ionisation energy of He^{+} is  19.6\times 10^{-18}J atom -1 The energy of the first stationary state (n=1) of Li^{2+} is

  • Option 1)

    8.82 × 10-17 J atom -1

  • Option 2)

    4.41× 10-16 J atom -1

  • Option 3)

    - 4.41× 10-17 J atom -1

  • Option 4)

    - 2.2× 10-15 J atom -1

 

Answers (2)

As we learnt in 

Total energy of elctron in nth orbit -

E_{n}= -13.6\: \frac{z^{2}}{n^{2}}eV

Where z is atomic number

-

 E_n = -E_o \times \frac{2^2}{1^2} = 19.6 \times 10^{-18}

E_o = \frac {19.6 \times 10^{-18}}{4}

E_1 for Li^{+2} = \frac {19.6 \times 10^{-18}}{4} \times 9

                    = -4.14 \times 10^{-17} J atm^{-1}

 


Option 1)

8.82 × 10-17 J atom -1

Incorrect

Option 2)

4.41× 10-16 J atom -1

Incorrect

Option 3)

- 4.41× 10-17 J atom -1

Correct

Option 4)

- 2.2× 10-15 J atom -1

Incorrect

Posted by

Vakul

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