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.
Can a single photon, emitted in the first and//or sixth line of the Lyman series, ionize:
{:a) another hydrogen atom in its ground state?
{:b) a copper atom in the Cu crystal?
The electron work function of Cu is `phi_(Cu)=7.44xx10^(-19) J`.

Answer

To solve the problem, we need to analyze the energy of the photons emitted in the first and sixth lines of the Lyman series and determine if they can ionize another hydrogen atom in its ground state and a copper atom in a Cu crystal.

### Step-by-Step Solution:

1. **Understanding the Lyman Series**:
The Lyman series corresponds to transitions of electrons in a hydrogen atom from higher energy levels (n ≥ 2) to the ground state (n = 1). The energy of the emitted photon can be calculated using the formula for the energy levels of hydrogen.

2. **Calculate the Energy for the First Line (n=2 to n=1)**:
- For n = 2:
\[ E_2 = -\frac{2.18 \times 10^{-8}}{2^2} = -\frac{2.18 \times 10^{-8}}{4} = -5.45 \times 10^{-9} \text{ J} \]
- For n = 1:
\[ E_1 = -2.18 \times 10^{-8} \text{ J} \]
- Energy of the photon emitted (E_photon) when an electron transitions from n=2 to n=1:
\[ E_{\text{photon}} = E_1 - E_2 = -2.18 \times 10^{-8} - (-5.45 \times 10^{-9}) = -2.18 \times 10^{-8} + 5.45 \times 10^{-9} = -1.635 \times 10^{-8} \text{ J} \]
- Thus, the energy of the photon for the first line is:
\[ E_{\text{photon}} = 1.635 \times 10^{-8} \text{ J} \]

3. **Calculate the Energy for the Sixth Line (n=7 to n=1)**:
- For n = 7:
\[ E_7 = -\frac{2.18 \times 10^{-8}}{7^2} = -\frac{2.18 \times 10^{-8}}{49} = -4.45 \times 10^{-10} \text{ J} \]
- Energy of the photon emitted when an electron transitions from n=7 to n=1:
\[ E_{\text{photon}} = E_1 - E_7 = -2.18 \times 10^{-8} - (-4.45 \times 10^{-10}) = -2.18 \times 10^{-8} + 4.45 \times 10^{-10} = -2.1355 \times 10^{-8} \text{ J} \]
- Thus, the energy of the photon for the sixth line is:
\[ E_{\text{photon}} = 2.1355 \times 10^{-8} \text{ J} \]

4. **Determine Ionization Energy of Hydrogen Atom**:
The ionization energy of a hydrogen atom in its ground state (n=1) is:
\[ E_{\text{ionization}} = -E_1 = 2.18 \times 10^{-8} \text{ J} \]

5. **Check if Photons Can Ionize Hydrogen Atom**:
- For the first line:
\[ E_{\text{photon}} = 1.635 \times 10^{-8} \text{ J} < 2.18 \times 10^{-8} \text{ J} \quad \text{(cannot ionize)} \]
- For the sixth line:
\[ E_{\text{photon}} = 2.1355 \times 10^{-8} \text{ J} < 2.18 \times 10^{-8} \text{ J} \quad \text{(cannot ionize)} \]

6. **Determine if Photons Can Ionize Copper Atom**:
The work function of copper is given as:
\[ \phi_{\text{Cu}} = 7.44 \times 10^{-19} \text{ J} \]
- For the first line:
\[ E_{\text{photon}} = 1.635 \times 10^{-8} \text{ J} > 7.44 \times 10^{-19} \text{ J} \quad \text{(can ionize)} \]
- For the sixth line:
\[ E_{\text{photon}} = 2.1355 \times 10^{-8} \text{ J} > 7.44 \times 10^{-19} \text{ J} \quad \text{(can ionize)} \]

### Final Answers:
- **a)** A single photon emitted in the first and/or sixth line of the Lyman series **cannot ionize** another hydrogen atom in its ground state.
- **b)** A single photon emitted in the first and/or sixth line of the Lyman series **can ionize** a copper atom in the Cu crystal.