Problem

(a) Determine the energy associated with the innermost orbit ($n=1$) of the hydrogen atom.
(b) Determine the energy associated with the second orbit ($n=2$) of the hydrogen atom.
(c) Calculate the energy required for a photon to raise an electron from the first to the second allowed orbit.

Data

• Energy Level ($n$): $1, 2$
• Planck’s constant ($h$): $6.626 \times 10^{-34} \, \mathrm{Js}$
• Electron mass ($m$): $9.109 \times 10^{-31} \, \mathrm{kg}$
• Coulomb’s constant ($k$): $8.988 \times 10^{9} \, \mathrm{Nm}^2/\mathrm{C}^2$
• Charge on electron ($e$): $1.602 \times 10^{-19} \, \mathrm{C}$

To Find:

1. Energy of the first orbit ($E_1$)
2. Energy of the second orbit ($E_2$)
3. Energy difference ($\Delta E$)

Prerequisite Concepts

1. Energy of the $n$-th Orbit:
   The energy of an electron in the $n$-th orbit of the hydrogen atom is given by:

$$E_n = -\frac{1}{n^2} \cdot \frac{2 \pi^2 m k^2 e^4}{h^2}$$

Where:

• $n$: Principal quantum number (orbit level)
• $m$: Mass of the electron
• $k$: Coulomb’s constant
• $e$: Charge of the electron
• $h$: Planck’s constant

2. Energy Difference ($\Delta E$):
   The energy required to excite an electron from the first orbit to the second orbit is:

$$\Delta E = E_2 - E_1$$

3. Energy Conversion:
   Convert energy from joules to electronvolts using:

$$1 \, \mathrm{eV} = 1.602 \times 10^{-19} \, \mathrm{J}$$

Solution

Part (a): Energy of the First Orbit ($E_1$)

Substitute the given values into the formula for $E_n$ with $n=1$:

$$E_1 = -\frac{1}{1^2} \cdot \frac{2 (3.14)^2 (9.109 \times 10^{-31}) (8.988 \times 10^{9})^2 (1.602 \times 10^{-19})^4}{(6.626 \times 10^{-34})^2}$$

Simplify:

$$E_1 = -2.17 \times 10^{-18} \, \mathrm{J}$$

Convert to electronvolts:

$$E_1 = \frac{-2.17 \times 10^{-18}}{1.602 \times 10^{-19}} = -13.6 \, \mathrm{eV}$$

Part (b): Energy of the Second Orbit ($E_2$)

Substitute $n=2$ into the formula for $E_n$:

$$E_2 = -\frac{1}{2^2} \cdot \frac{2 (3.14)^2 (9.109 \times 10^{-31}) (8.988 \times 10^{9})^2 (1.602 \times 10^{-19})^4}{(6.626 \times 10^{-34})^2}$$

Simplify:

$$E_2 = -0.54 \times 10^{-18} \, \mathrm{J}$$

Convert to electronvolts:

$$E_2 = \frac{-0.54 \times 10^{-18}}{1.602 \times 10^{-19}} = -3.4 \, \mathrm{eV}$$

Part (c): Energy Difference ($\Delta E$)

$$\Delta E = E_2 - E_1$$

Substitute the values of $E_2$ and $E_1$:

$$\Delta E = -3.4 \, \mathrm{eV} - (-13.6 \, \mathrm{eV})$$ $$\Delta E = 10.2 \, \mathrm{eV}$$

Answer

1. Energy of the first orbit:

$$E_1 = -13.6 \, \mathrm{eV}$$

2. Energy of the second orbit:

$$E_2 = -3.4 \, \mathrm{eV}$$

3. Energy required to excite an electron from $n=1$ to $n=2$:

$$\Delta E = 10.2 \, \mathrm{eV}$$

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