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The ionization energy of the electron in...

The ionization energy of the electron in the lowest orbit of hydrogen atom is 13.6 eV. The energies required in eV to remove an electron from three lowest energy orbits of hydrogen atom respectively are

A

`13.6, 6.8, 8.4`

B

`13.6,10.2, 3.4`

C

`13.6, 27.2, 40.8`

D

`13.6, 3.4, 1.51`

Text Solution

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The correct Answer is:
To find the energies required to remove an electron from the three lowest energy orbits of a hydrogen atom, we will use the formula for the energy of an electron in the nth orbit according to Bohr's model: \[ E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \] where: - \( E_n \) is the energy of the electron in the nth orbit, - \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)), - \( n \) is the principal quantum number (orbit number). ### Step 1: Calculate the energy for the first orbit (n = 1) Using the formula: \[ E_1 = -\frac{13.6 \, \text{eV} \cdot 1^2}{1^2} = -13.6 \, \text{eV} \] The energy required to remove the electron from the first orbit to infinity is: \[ \Delta E_1 = E_{\infty} - E_1 = 0 - (-13.6) = 13.6 \, \text{eV} \] ### Step 2: Calculate the energy for the second orbit (n = 2) Using the formula: \[ E_2 = -\frac{13.6 \, \text{eV} \cdot 1^2}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \] The energy required to remove the electron from the second orbit to infinity is: \[ \Delta E_2 = E_{\infty} - E_2 = 0 - (-3.4) = 3.4 \, \text{eV} \] ### Step 3: Calculate the energy for the third orbit (n = 3) Using the formula: \[ E_3 = -\frac{13.6 \, \text{eV} \cdot 1^2}{3^2} = -\frac{13.6 \, \text{eV}}{9} \approx -1.511 \, \text{eV} \] The energy required to remove the electron from the third orbit to infinity is: \[ \Delta E_3 = E_{\infty} - E_3 = 0 - (-1.511) = 1.511 \, \text{eV} \] ### Final Answers The energies required to remove an electron from the three lowest energy orbits of the hydrogen atom are: 1. From the first orbit (n=1): **13.6 eV** 2. From the second orbit (n=2): **3.4 eV** 3. From the third orbit (n=3): **1.511 eV** ### Summary - Energy from n=1: 13.6 eV - Energy from n=2: 3.4 eV - Energy from n=3: 1.511 eV

To find the energies required to remove an electron from the three lowest energy orbits of a hydrogen atom, we will use the formula for the energy of an electron in the nth orbit according to Bohr's model: \[ E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \] where: - \( E_n \) is the energy of the electron in the nth orbit, - \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)), - \( n \) is the principal quantum number (orbit number). ...
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