The ratio of the energy of the electron in the ground state of hydrogen to the electron in the first excited state of Be^{3+} is 

  • Option 1)

    1:4

  • Option 2)

    1:8

  • Option 3)

    1:16

  • Option 4)

    16:1

 

Answers (1)
P Prateek Shrivastava

As learnt in

Total energy of elctron in nth orbit -

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

Where z is atomic number

-

 

 We know that total energy of electron in nth orbit, E_{n} = -13.6 \frac{z^{2}}{n^{2}}eV

E1 (ground state of hydrogen) = -13.6 \times \frac{1}{1}eV

E2 (first excited state of Be 3+) = -13.6 \times \frac{4\times 4}{2\times 2}

        (n = 2, z = 4)

                                                  = -13.6 x 4 eV

\therefore E_{1} : E_{2} = 1 : 4


Option 1)

1:4

This option is correct

Option 2)

1:8

This option is incorrect

Option 3)

1:16

This option is incorrect

Option 4)

16:1

This option is incorrect

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