If 13.6 eV energy is required to ionize the hydrogen atom, then the energy required to remove an electron from n = 2 is

  • Option 1)

    10.2 eV

  • Option 2)

    0 eV

  • Option 3)

    3.4 eV

  • Option 4)

    6.8 eV

 

Answers (1)
P perimeter

As we learnt in

Ionization energy -

E_{ion }= E_{\infty }-E_{n }

=13.6\frac{Z^{2}}{n^{2}}

 

- wherein

Energy required to move an electron from ground state to n= \infty

 

 \Delta E =E_{0}z^{2}\left(\frac{1}{n_{i}^{2}}-\frac{1}{n_{f}^{2}} \right )

For an electron to remove from 

n = 2

ni = 2

n_{f}\rightarrow \infty

E_{0}=13.6\ eV     &    z = 1

\therefore\ \; \Delta E=13.6\left(\frac{1}{4} \right )=3.4\ eV

Correct option is 3.


Option 1)

10.2 eV

This is an incorrect option.

Option 2)

0 eV

This is an incorrect option.

Option 3)

3.4 eV

This is the correct option.

Option 4)

6.8 eV

This is an incorrect option.

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