the wavelength of spectral line in the lyman series of a H-atom is `1028 Å`. If instead of hydrogen, we consider deuterium then shift in the wavelength of this line be `(m_p=1860m_e)`

Options

A. `1027.7Å`
B. `1036Å`
C. `1028Å`
D. `1021Å`

Explanation

To solve the problem of finding the shift in the wavelength of the spectral line in the Lyman series when considering deuterium instead of hydrogen, we will follow these steps:

### Step 1: Understand the Concept of Reduced Mass
The wavelength of spectral lines in hydrogen and deuterium depends on the reduced mass of the system. The reduced mass \( \mu \) is given by the formula:
\[ \mu = \frac{m_e \cdot m_p}{m_e + m_p} \]
where \( m_e \) is the mass of the electron and \( m_p \) is the mass of the proton.

### Step 2: Calculate the Reduced Mass for Hydrogen
For hydrogen, the mass of the proton is much larger than the mass of the electron, so we can approximate:
\[ \mu_H \approx \frac{m_e \cdot m_p}{m_p} = m_e \]

### Step 3: Calculate the Reduced Mass for Deuterium
In deuterium, there is one proton and one neutron in the nucleus. The mass of deuterium is approximately twice the mass of the proton:
\[ m_{D} \approx 2m_p \]
Thus, the reduced mass for deuterium is:
\[ \mu_D = \frac{m_e \cdot 2m_p}{m_e + 2m_p} \]

### Step 4: Set Up the Ratio of Wavelengths
The wavelengths of the spectral lines are inversely proportional to the reduced mass:
\[ \frac{\lambda_D}{\lambda_H} = \frac{\mu_H}{\mu_D} \]
Rearranging gives:
\[ \lambda_D = \lambda_H \cdot \frac{\mu_D}{\mu_H} \]

### Step 5: Substitute the Reduced Mass Values
Substituting the expressions for reduced mass:
\[ \lambda_D = \lambda_H \cdot \frac{\frac{m_e \cdot 2m_p}{m_e + 2m_p}}{m_e} \]
This simplifies to:
\[ \lambda_D = \lambda_H \cdot \frac{2m_p}{m_e + 2m_p} \]

### Step 6: Substitute Known Values
Given \( \lambda_H = 1028 \, \text{Å} \) and \( m_p = 1860m_e \):
\[ \lambda_D = 1028 \cdot \frac{2 \cdot 1860m_e}{m_e + 2 \cdot 1860m_e} \]
This simplifies to:
\[ \lambda_D = 1028 \cdot \frac{3720m_e}{3721m_e} \]
The \( m_e \) cancels out:
\[ \lambda_D = 1028 \cdot \frac{3720}{3721} \]

### Step 7: Calculate the Final Wavelength
Now, calculate:
\[ \lambda_D \approx 1028 \cdot (1 - \frac{1}{3721}) \approx 1028 \cdot (1 - 0.000268) \approx 1028 - 0.276 \approx 1027.724 \, \text{Å} \]
Rounding gives:
\[ \lambda_D \approx 1027.7 \, \text{Å} \]

### Conclusion
Thus, the shift in the wavelength of the spectral line in the Lyman series when considering deuterium instead of hydrogen is approximately:
\[ \lambda_D \approx 1027.7 \, \text{Å} \]