The orbital speed of the electron in the ground state of hydrogen is `v`. What will be its orbital speed when it is excited to the enrgy state `-3.4 eV`?

Options

A. `2v`
B. `(v)/(2)`
C. `(v)/(4)`
D. `(v)/(8)`

Explanation

To solve the problem of finding the orbital speed of an electron in an excited state of hydrogen, we can follow these steps:

### Step 1: Understand the relationship between energy levels and speed
The energy levels of an electron in a hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number (1 for ground state, 2 for first excited state, etc.).

### Step 2: Identify the ground state energy
For the ground state (n=1), the energy is:
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]

### Step 3: Identify the excited state energy
The excited state energy given in the problem is:
\[ E = -3.4 \, \text{eV} \]

### Step 4: Determine the principal quantum number for the excited state
Using the energy formula, we can find \( n \) for the excited state:
\[ -3.4 = -\frac{13.6}{n^2} \]
Rearranging gives:
\[ n^2 = \frac{13.6}{3.4} = 4 \quad \Rightarrow \quad n = 2 \]

### Step 5: Relate the orbital speed to the principal quantum number
The orbital speed \( v \) of an electron in a hydrogen atom is inversely proportional to the square root of the principal quantum number:
\[ v_n \propto \frac{1}{n} \]
Thus, the speed in the ground state (n=1) is \( v_1 \), and the speed in the excited state (n=2) is \( v_2 \):
\[ \frac{v_2}{v_1} = \frac{1}{n_2} = \frac{1}{2} \]

### Step 6: Calculate the speed in the excited state
If the speed in the ground state is \( v \), then:
\[ v_2 = \frac{v}{2} \]

### Final Answer
The orbital speed of the electron when it is excited to the energy state of -3.4 eV is:
\[ v_2 = \frac{v}{2} \]

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