The energy of stable states of the hydrogen atom is given by `E_(n)=-2.18xx10^(-8)//n^(2)[J]` where n denotes the principal quantum number.
Calculate the de Broglie wavelength of the electrons emitted from a copper crystal when irradiated by photons from the first line and the sixth line of the Lyman series.
`h=6.6256xx10^(-34) J s," "m_(e)=9.1091xx10^(31) kg," "c=2.99792xx10^(8) m s^(-1)`

Answer

To solve the problem of calculating the de Broglie wavelength of the electrons emitted from a copper crystal when irradiated by photons from the first line and the sixth line of the Lyman series, we will follow these steps:

### Step 1: Calculate the energy of the photons from the Lyman series.

The energy of the photons emitted in the Lyman series can be calculated using the formula for the energy levels of the hydrogen atom:

\[ E_n = -\frac{2.18 \times 10^{-8}}{n^2} \text{ J} \]

For the Lyman series, the transitions are from higher energy levels (n = 2, 3, 4, ...) to n = 1.

- **First line (n = 2 to n = 1):**
\[ E_1 = -\frac{2.18 \times 10^{-8}}{1^2} = -2.18 \times 10^{-8} \text{ J} \]
\[ E_2 = -\frac{2.18 \times 10^{-8}}{2^2} = -\frac{2.18 \times 10^{-8}}{4} = -5.45 \times 10^{-9} \text{ J} \]
\[ \Delta E_{1 \to 2} = E_1 - E_2 = (-2.18 \times 10^{-8}) - (-5.45 \times 10^{-9}) = -2.18 \times 10^{-8} + 5.45 \times 10^{-9} = -1.635 \times 10^{-8} \text{ J} \]

- **Sixth line (n = 7 to n = 1):**
\[ E_7 = -\frac{2.18 \times 10^{-8}}{7^2} = -\frac{2.18 \times 10^{-8}}{49} = -4.45 \times 10^{-10} \text{ J} \]
\[ \Delta E_{1 \to 7} = E_1 - E_7 = (-2.18 \times 10^{-8}) - (-4.45 \times 10^{-10}) = -2.18 \times 10^{-8} + 4.45 \times 10^{-10} = -2.1355 \times 10^{-8} \text{ J} \]

### Step 2: Calculate the kinetic energy of the emitted electrons.

The energy of the emitted electrons is equal to the energy of the photons minus the work function (which we will assume is negligible for this calculation). Thus, we can use the energy calculated above directly.

### Step 3: Calculate the de Broglie wavelength of the emitted electrons.

The de Broglie wavelength (\(\lambda\)) is given by the formula:

\[ \lambda = \frac{h}{p} \]

where \(p\) is the momentum of the electron, which can be calculated using its kinetic energy (\(KE\)):

\[ KE = \frac{p^2}{2m_e} \implies p = \sqrt{2m_e \cdot KE} \]

Substituting this into the de Broglie wavelength formula gives:

\[ \lambda = \frac{h}{\sqrt{2m_e \cdot KE}} \]

### Step 4: Calculate for both lines.

1. **For the first line:**
- \(KE = 1.635 \times 10^{-8} \text{ J}\)
- Substitute \(h = 6.6256 \times 10^{-34} \text{ J s}\) and \(m_e = 9.1091 \times 10^{-31} \text{ kg}\):

\[ \lambda_1 = \frac{6.6256 \times 10^{-34}}{\sqrt{2 \cdot 9.1091 \times 10^{-31} \cdot 1.635 \times 10^{-8}}} \]

2. **For the sixth line:**
- \(KE = 2.1355 \times 10^{-8} \text{ J}\)

\[ \lambda_2 = \frac{6.6256 \times 10^{-34}}{\sqrt{2 \cdot 9.1091 \times 10^{-31} \cdot 2.1355 \times 10^{-8}}} \]

### Step 5: Final calculations.

Now, compute the values for \(\lambda_1\) and \(\lambda_2\).