What is the energy required to move the electron from the ground state of H atom to the first excited state? Given that the ground state energy of H atom is 13.6 eV and that the energy E $_{n}$ of an electron in n $^{t h}$ orbital of an atom or ion of atomic number Z is, given by the equation $E_{n} = (13.6 Z^{2} / n^{2})$

13.6 eV

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3.4 eV

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10.2 eV

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-10.2 eV

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Solution

The correct option is A 10.2 eV
Energy, $E_{n} = - \frac{13.6}{n^{2}} Z^{2}$

The energy required to move the electron from ground state (n =1) of H to first excited state (n= 2) is

$Δ E = E_{2} - E_{1} = - 13.6 (\frac{1}{2^{2}} - \frac{1}{1^{2}})$

$Δ E = 13.6 \times \frac{3}{4} = 10.2 e V$

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What is the energy required to move the electron from the ground state of H atom to the first excited state? Given that the ground state energy of H atom is 13.6 eV and that the energy En of an electron in nth orbital of an atom or ion of atomic number Z is, given by the equation En=(13.6Z2/n2)

The correct option is A 10.2 eVEnergy, En=−13.6n2Z2The energy required to move the electron from ground state (n =1) of H to first excited state (n= 2) is ΔE=E2−E1=−13.6(122−112)ΔE=13.6×34=10.2 eV

The correct option is A 10.2 eV
Energy, $E_{n} = - \frac{13.6}{n^{2}} Z^{2}$

The energy required to move the electron from ground state (n =1) of H to first excited state (n= 2) is

$Δ E = E_{2} - E_{1} = - 13.6 (\frac{1}{2^{2}} - \frac{1}{1^{2}})$

$Δ E = 13.6 \times \frac{3}{4} = 10.2 e V$