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A beam of 13.0 eV electrons is used to b...

A beam of `13.0 eV` electrons is used to bombard gaseous hydrogen. The series obtained in emission spectra is // are

A

Lyman series

B

Balmer series

C

Brackett series

D

All of these

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To solve the problem of determining the emission series obtained when a beam of 13.0 eV electrons bombards gaseous hydrogen, we will follow these steps: ### Step 1: Understand the Energy Levels of Hydrogen The energy levels of the hydrogen atom can be described by the formula: \[ E_n = -\frac{13.6 \text{ eV}}{n^2} \] where \( n \) is the principal quantum number (n = 1, 2, 3, ...). ### Step 2: Calculate the Energy of the First Few Levels - For \( n = 1 \): \[ E_1 = -\frac{13.6 \text{ eV}}{1^2} = -13.6 \text{ eV} \] - For \( n = 2 \): \[ E_2 = -\frac{13.6 \text{ eV}}{2^2} = -3.4 \text{ eV} \] - For \( n = 3 \): \[ E_3 = -\frac{13.6 \text{ eV}}{3^2} \approx -1.51 \text{ eV} \] - For \( n = 4 \): \[ E_4 = -\frac{13.6 \text{ eV}}{4^2} = -0.85 \text{ eV} \] - For \( n = 5 \): \[ E_5 = -\frac{13.6 \text{ eV}}{5^2} \approx -0.544 \text{ eV} \] ### Step 3: Determine the Maximum Excitation Level The energy provided to the electrons is 13.0 eV. To find out how far the electrons can be excited, we need to calculate the energy difference between the ground state and the excited state: \[ \text{Energy provided} = 13.0 \text{ eV} - (-13.6 \text{ eV}) = 13.0 + 13.6 = 26.6 \text{ eV} \] This means the electrons can be excited to a level where the total energy is less than or equal to 0 eV. ### Step 4: Find the Maximum n Value We need to find the maximum \( n \) such that: \[ E_n \geq -13.0 \text{ eV} \] From the energy level formula: \[ -\frac{13.6}{n^2} \geq -13.0 \] This simplifies to: \[ \frac{13.6}{n^2} \leq 13.0 \implies n^2 \geq \frac{13.6}{13.0} \approx 1.046 \] Thus, \( n \) can be 1 or higher. ### Step 5: Identify Possible Emission Series The electrons can de-excite from the maximum level they reach. The maximum level can be calculated as follows: - If the electrons reach \( n = 4 \) (as determined from energy levels), they can de-excite to: - \( n = 3 \) (Paschen series) - \( n = 2 \) (Balmer series) - \( n = 1 \) (Lyman series) ### Conclusion The possible emission series are: - Lyman series (from \( n=2 \) to \( n=1 \)) - Balmer series (from \( n=3 \) to \( n=2 \)) - Paschen series (from \( n=4 \) to \( n=3 \)) ### Final Answer The series obtained in the emission spectra are the **Lyman series, Balmer series, and Paschen series**. ---

To solve the problem of determining the emission series obtained when a beam of 13.0 eV electrons bombards gaseous hydrogen, we will follow these steps: ### Step 1: Understand the Energy Levels of Hydrogen The energy levels of the hydrogen atom can be described by the formula: \[ E_n = -\frac{13.6 \text{ eV}}{n^2} \] where \( n \) is the principal quantum number (n = 1, 2, 3, ...). ...
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