# 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

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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