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 can follow these steps:

### Step 1: Understand the relationship between the wavelengths
The relationship between the wavelengths of hydrogen and deuterium can be expressed as:
\[ \frac{\lambda_D}{\lambda_H} = 1 - \frac{m_e}{2m_p} \]
where:
- \(\lambda_D\) is the wavelength for deuterium,
- \(\lambda_H\) is the wavelength for hydrogen,
- \(m_e\) is the mass of the electron,
- \(m_p\) is the mass of the proton.

### Step 2: Substitute the known values
Given:
- \(\lambda_H = 1028 \, \text{Å}\)
- \(m_p = 1860 \, m_e\)

We can substitute \(m_p\) into the equation:
\[ \frac{\lambda_D}{\lambda_H} = 1 - \frac{m_e}{2 \times 1860 m_e} \]

### Step 3: Simplify the equation
The mass of the electron (\(m_e\)) cancels out:
\[ \frac{\lambda_D}{\lambda_H} = 1 - \frac{1}{3720} \]
Now, calculate \(1 - \frac{1}{3720}\):
\[ 1 - \frac{1}{3720} = \frac{3720 - 1}{3720} = \frac{3719}{3720} \]

### Step 4: Calculate the wavelength of deuterium
Now, substitute \(\lambda_H\) into the equation:
\[ \lambda_D = \lambda_H \times \left(1 - \frac{1}{3720}\right) = 1028 \, \text{Å} \times \frac{3719}{3720} \]

### Step 5: Perform the calculation
Calculating \(\lambda_D\):
\[ \lambda_D = 1028 \times \frac{3719}{3720} \]
\[ \lambda_D \approx 1028 \times 0.99973 \approx 1027.7 \, \text{Å} \]

### Step 6: Determine the shift in wavelength
The shift in wavelength (\(\Delta \lambda\)) is given by:
\[ \Delta \lambda = \lambda_H - \lambda_D \]
Substituting the values:
\[ \Delta \lambda = 1028 \, \text{Å} - 1027.7 \, \text{Å} = 0.3 \, \text{Å} \]

### Final Answer
The shift in the wavelength of the spectral line in the Lyman series when considering deuterium instead of hydrogen is approximately \(0.3 \, \text{Å}\).

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