According to de - Broglie , the de - Broglie wavelength for electron in an orbit of hydrogen atom is `10^(-9) m`. The principle quantum number for this electron is

Explanation

To find the principal quantum number for an electron in a hydrogen atom with a given de Broglie wavelength, we can follow these steps:

### Step 1: Understand the given information
We are given the de Broglie wavelength (λ) of the electron in the hydrogen atom, which is \(10^{-9}\) m.

### Step 2: Use the de Broglie wavelength formula
The de Broglie wavelength 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 its velocity.

### Step 3: Relate momentum to the principal quantum number
According to Bohr's model of the hydrogen atom, the momentum \(mv\) can also be expressed in terms of the principal quantum number \(n\):
\[ mv = \frac{nh}{2\pi r} \]
where \(r\) is the radius of the orbit.

### Step 4: Substitute the momentum into the de Broglie wavelength formula
Substituting \(mv\) in the de Broglie wavelength formula, we have:
\[ \lambda = \frac{h}{\frac{nh}{2\pi r}} = \frac{2\pi r}{n} \]

### Step 5: Rearrange the equation to find \(n\)
Rearranging gives:
\[ n = \frac{2\pi r}{\lambda} \]

### Step 6: Use the radius of the first orbit
The radius of the first orbit in a hydrogen atom is given as:
\[ r_1 = 0.053 \text{ nm} = 0.053 \times 10^{-9} \text{ m} \]

### Step 7: Substitute the values into the equation
Now substituting the values of \(r\) and \(\lambda\):
\[ n = \frac{2\pi (0.053 \times 10^{-9})}{10^{-9}} \]

### Step 8: Simplify the equation
This simplifies to:
\[ n = 2\pi \times 0.053 \]

### Step 9: Calculate the value of \(n\)
Calculating this gives:
\[ n \approx 0.333 \times 2\pi \approx 1 \]

### Step 10: Final answer
Thus, the principal quantum number \(n\) for the electron in the hydrogen atom is approximately 1.

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