The binding energy of the electron in the lowest orbit of the hydrogen atom is 13.6 eV. The magnitudes energies from three lowest orbits of the hydrogen are

Question

The binding energy of the electron in the lowest orbit of the hydrogen atom is 13.6 eV. The magnitudes energies from three lowest orbits of the hydrogen are (A) 13.6, 6.8, 8.4 eV (B) 13.6, 10.2, 3.4 eV (C) 13.6, 27.2, 40.8 eV (D) 13.6, 3.4, 1.5 eV

Tardigrade Admin · Accepted Answer

En=-13.6(Z2/n2) I orbit for (H) Z=1,n=1 En=-13.6eV II orbit for (H) Z=1,n=2 En=-(13.6/4)=-3.4eV III orbit for (H) En=-(13.6/9)=-1.51eV

Q. The binding energy of the electron in the lowest orbit of the hydrogen atom is 13.6 eV. The magnitudes energies from three lowest orbits of the hydrogen are

13.6, 6.8, 8.4 eV

13.6, 10.2, 3.4 eV

13.6, 27.2, 40.8 eV

13.6, 3.4, 1.5 eV

$E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}}$

I orbit for (H)

$Z = 1, n = 1$

$E_{n} = - 13.6 e V$

II orbit for (H)

$Z = 1, n = 2$

$E_{n} = - \frac{13.6}{4} = - 3.4 e V$

III orbit for (H)

$E_{n} = - \frac{13.6}{9} = - 1.51 e V$

Q. The binding energy of the electron in the lowest orbit of the hydrogen atom is 13.6 eV. The magnitudes energies from three lowest orbits of the hydrogen are

13.6, 6.8, 8.4 eV

13.6, 10.2, 3.4 eV

13.6, 27.2, 40.8 eV

13.6, 3.4, 1.5 eV

En​=−13.6n2Z2​ I orbit for (H) Z=1,n=1 En​=−13.6eV II orbit for (H) Z=1,n=2 En​=−413.6​=−3.4eV III orbit for (H) En​=−913.6​=−1.51eV

$E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}}$

I orbit for (H)

$Z = 1, n = 1$

$E_{n} = - 13.6 e V$

II orbit for (H)

$Z = 1, n = 2$

$E_{n} = - \frac{13.6}{4} = - 3.4 e V$

III orbit for (H)

$E_{n} = - \frac{13.6}{9} = - 1.51 e V$