Calculate the number of waves made by a Bohr electron in one complete revolution in nth orbit of `He^(+)` ion, if ratio of de-Broglie wavelength associated with electron moving in `n^(th)` orbit and `2^(nd)` orbit is 2.0 :-

Answer

To solve the problem of calculating the number of waves made by a Bohr electron in one complete revolution in the nth orbit of the He⁺ ion, we can follow these steps:

### Step 1: Understand the de Broglie wavelength
The de Broglie wavelength (λ) of an electron in a circular orbit is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where \(h\) is Planck's constant, \(m\) is the mass of the electron, and \(v\) is the velocity of the electron.

### Step 2: Relate de Broglie wavelength to the orbit
In the Bohr model, the circumference of the nth orbit is equal to an integer multiple of the de Broglie wavelength:
\[ 2\pi r_n = n\lambda_n \]
where \(r_n\) is the radius of the nth orbit and \(n\) is the principal quantum number.

### Step 3: Set up the ratio of wavelengths
According to the problem, the ratio of the de Broglie wavelengths in the nth orbit and the 2nd orbit is given as:
\[ \frac{\lambda_n}{\lambda_2} = 2.0 \]
This implies:
\[ \lambda_n = 2\lambda_2 \]

### Step 4: Substitute the relationship into the wavelength equation
From the relationship established in Step 2, we can express the wavelengths in terms of the orbits:
\[ \lambda_n = \frac{2\pi r_n}{n} \]
\[ \lambda_2 = \frac{2\pi r_2}{2} \]
Substituting the ratio:
\[ \frac{2\pi r_n}{n} = 2 \cdot \frac{2\pi r_2}{2} \]
This simplifies to:
\[ \frac{r_n}{n} = 2 \cdot \frac{r_2}{2} \]
Thus:
\[ r_n = n \cdot r_2 \]

### Step 5: Calculate the number of waves in the nth orbit
The number of waves made by the electron in one complete revolution in the nth orbit can be calculated as:
\[ \text{Number of waves} = \frac{\text{Circumference of the orbit}}{\text{Wavelength}} = \frac{2\pi r_n}{\lambda_n} \]
Substituting \(r_n\) and \(\lambda_n\):
\[ \text{Number of waves} = \frac{2\pi (n \cdot r_2)}{2\lambda_2} = \frac{n \cdot 2\pi r_2}{2\lambda_2} \]
From the relationship \(2\pi r_2 = 2\lambda_2\), we can simplify:
\[ \text{Number of waves} = n \]

### Step 6: Conclusion
Since we have established that the number of waves made by the electron in the nth orbit is equal to \(n\), and given that the ratio of wavelengths indicates that the electron makes 2 waves for every wave in the 2nd orbit, we can conclude:
\[ \text{Total number of waves} = 2n \]

### Final Answer
For \(n = 2\), the total number of waves made by the electron in one complete revolution in the nth orbit of the He⁺ ion is:
\[ \text{Total number of waves} = 4 \]