If `lamda_1` and `lamda_2` denote the wavelength of de Broglie waves for electrons in the first and second Bohr orbits in a hydrogen atom, the `(lamda_1)/(lamda_2)` is equal to

Options

A. `(2)/(1)`
B. `(1)/(2)`
C. `(1)/(4)`
D. `(4)/(1)`

Explanation

To solve the problem of finding the ratio of the de Broglie wavelengths \( \frac{\lambda_1}{\lambda_2} \) for electrons in the first and second Bohr orbits of a hydrogen atom, we can follow these steps:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength \( \lambda \) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

### Step 2: Relate momentum to velocity
For an electron, the momentum \( p \) can be expressed as:
\[ p = mv \]
where \( m \) is the mass of the electron and \( v \) is its velocity. Therefore, we can rewrite the de Broglie wavelength as:
\[ \lambda = \frac{h}{mv} \]

### Step 3: Establish the relationship between wavelengths in different orbits
Let \( \lambda_1 \) be the wavelength of the electron in the first Bohr orbit and \( \lambda_2 \) be the wavelength in the second Bohr orbit. From the previous formula, we can say:
\[ \lambda_1 = \frac{h}{m v_1} \quad \text{and} \quad \lambda_2 = \frac{h}{m v_2} \]
Taking the ratio of the two wavelengths gives:
\[ \frac{\lambda_1}{\lambda_2} = \frac{v_2}{v_1} \]

### Step 4: Use Bohr's postulate for velocity
According to Bohr's model, the velocity of an electron in the nth orbit is inversely proportional to the principal quantum number \( n \):
\[ v \propto \frac{1}{n} \]
Thus, we can express the velocities in terms of the principal quantum numbers:
\[ v_1 \propto \frac{1}{n_1} \quad \text{and} \quad v_2 \propto \frac{1}{n_2} \]

### Step 5: Substitute the velocities into the wavelength ratio
Substituting the expressions for \( v_1 \) and \( v_2 \) into the wavelength ratio, we have:
\[ \frac{\lambda_1}{\lambda_2} = \frac{v_2}{v_1} = \frac{\frac{1}{n_2}}{\frac{1}{n_1}} = \frac{n_1}{n_2} \]

### Step 6: Substitute the principal quantum numbers
For the first Bohr orbit, \( n_1 = 1 \) and for the second Bohr orbit, \( n_2 = 2 \). Therefore:
\[ \frac{\lambda_1}{\lambda_2} = \frac{n_1}{n_2} = \frac{1}{2} \]

### Conclusion
Thus, the ratio of the de Broglie wavelengths for electrons in the first and second Bohr orbits is:
\[ \frac{\lambda_1}{\lambda_2} = \frac{1}{2} \]

### Final Answer
The correct answer is \( \frac{1}{2} \).

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