An electron in a hydrogen atom in its ground state absorbs 1.5 times as much energy as the minimum required for it to escape from the atom. What is the velocity of the emitted electron?

Options

A. `1.54xx10^(6)` m/s
B. `1.54xx10^(8)` m/s
C. `1.54xx10^(3)` m/s
D. `1.54xx10^(4)` m/s

Answer

To solve the problem of finding the velocity of the emitted electron from a hydrogen atom after it absorbs energy, we can follow these steps:

### Step 1: Determine the Ionization Energy
The ionization energy (IE) of a hydrogen atom in its ground state is given by the formula:
\[ IE = -\frac{13.6 \, \text{eV}}{n^2} \]
For hydrogen, \( n = 1 \):
\[ IE = -\frac{13.6 \, \text{eV}}{1^2} = 13.6 \, \text{eV} \]

### Step 2: Calculate the Energy Absorbed
The problem states that the electron absorbs 1.5 times the ionization energy:
\[ \text{Energy absorbed} = 1.5 \times IE = 1.5 \times 13.6 \, \text{eV} = 20.4 \, \text{eV} \]

### Step 3: Calculate the Total Energy After Absorption
The total energy of the electron after absorbing this energy is the sum of the ionization energy and the absorbed energy:
\[ \text{Total energy} = \text{Energy of ground state} + \text{Energy absorbed} \]
The energy of the ground state is \(-13.6 \, \text{eV}\):
\[ \text{Total energy} = -13.6 \, \text{eV} + 20.4 \, \text{eV} = 6.8 \, \text{eV} \]

### Step 4: Convert Energy to Joules
To find the kinetic energy in joules, we convert electron volts to joules using the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[ \text{Kinetic energy (KE)} = 6.8 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.088 \times 10^{-18} \, \text{J} \]

### Step 5: Use the Kinetic Energy Formula
The kinetic energy of the emitted electron can be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
Where \( m \) is the mass of the electron (\(9.1 \times 10^{-31} \, \text{kg}\)) and \( v \) is the velocity. Rearranging for \( v \):
\[ v = \sqrt{\frac{2 \times KE}{m}} \]

### Step 6: Substitute Values and Calculate Velocity
Substituting the values into the equation:
\[ v = \sqrt{\frac{2 \times 1.088 \times 10^{-18} \, \text{J}}{9.1 \times 10^{-31} \, \text{kg}}} \]
Calculating the value inside the square root:
\[ v = \sqrt{\frac{2.176 \times 10^{-18}}{9.1 \times 10^{-31}}} \]
\[ v = \sqrt{2.39 \times 10^{12}} \approx 1.55 \times 10^{6} \, \text{m/s} \]

### Final Answer
The velocity of the emitted electron is approximately:
\[ v \approx 1.54 \times 10^{6} \, \text{m/s} \]

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