Energy of the electron in nth orbit of hydrogen atom is given by En-13-6-n2 e V- The amount of energy needed to transfer electron from first orbit to third orbit is

The correct option is C 12.09 eVUsing E= 13.6/n^2 eV For n = 1, E_1 = 13.6/1^2 = 13.6 eV and for n = 3 E_3 = 13.6/3^2 = 1.51 eV So required energy =E_3 ...

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Standard XII

Physics

Energy Levels

Energy of the...

Question

Energy of the electron in $n^{t h}$ orbit of hydrogen atom is given by $E_{n} = - \frac{13.6}{n^{2}} e V$ . The amount of energy needed to
transfer electron from first orbit to third orbit is

13.6 eV

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3.4 eV

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12.09 eV

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1.51 eV

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Solution

The correct option is C 12.09 eV
Using $E = - \frac{13.6}{n^{2}} e V$
For n = 1, $E_{1} = \frac{- 13.6}{1^{2}} = - 13.6 e V$ and for n = 3 $E_{3} = - \frac{13.6}{3^{2}} = - 1.51 e V$
So required energy $= E_{3} - E_{1} = - 1.51 - (- 13.6) = 12.09 e V$

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Energy of the electron in nth orbit of hydrogen atom is given by En=−13.6n2eV. The amount of energy needed to transfer electron from first orbit to third orbit is

The correct option is C 12.09 eVUsing E=−13.6n2eV For n = 1, E1=−13.612=−13.6eV and for n = 3 E3=−13.632=−1.51eV So required energy =E3−E1=−1.51−(−13.6)=12.09eV

Energy of the electron in $n^{t h}$ orbit of hydrogen atom is given by $E_{n} = - \frac{13.6}{n^{2}} e V$ . The amount of energy needed to
transfer electron from first orbit to third orbit is

The correct option is C 12.09 eV
Using $E = - \frac{13.6}{n^{2}} e V$
For n = 1, $E_{1} = \frac{- 13.6}{1^{2}} = - 13.6 e V$ and for n = 3 $E_{3} = - \frac{13.6}{3^{2}} = - 1.51 e V$
So required energy $= E_{3} - E_{1} = - 1.51 - (- 13.6) = 12.09 e V$