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 the excited state of hydrogen when it is at an energy level of -3.4 eV, we can follow these steps:

### Step 1: Understand the energy levels of hydrogen
The energy of an electron in the nth energy level of a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

### Step 2: Determine the principal quantum number for -3.4 eV
We need to find \( n \) when the energy \( E_n = -3.4 \, \text{eV} \). Setting the equation for energy equal to -3.4 eV, we have:
\[ -3.4 = -\frac{13.6}{n^2} \]
By rearranging this equation, we can solve for \( n^2 \):
\[ n^2 = \frac{13.6}{3.4} \]
Calculating this gives:
\[ n^2 = 4 \implies n = 2 \]

### Step 3: Relate the speed of the electron to its energy level
The orbital speed \( v_n \) of an electron in the nth energy level is proportional to \( \frac{Z}{n} \), where \( Z \) is the atomic number. For hydrogen, \( Z = 1 \). Therefore, the speed can be expressed as:
\[ v_n \propto \frac{Z}{n} = \frac{1}{n} \]
Thus, for \( n = 1 \) (ground state), the speed is:
\[ v_1 = v \]
For \( n = 2 \) (excited state), the speed is:
\[ v_2 = \frac{v}{2} \]

### Step 4: Conclusion
The orbital speed of the electron when it is excited to the energy state of -3.4 eV (which corresponds to \( n = 2 \)) is:
\[ v_2 = \frac{v}{2} \]

### Final Answer
The orbital speed of the electron in the excited state is \( \frac{v}{2} \).

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