An electron is moving in `3^(rd)` orbit of Hydrogen series and radius of first orbit is x then

Options

A. de-Broglie wavelength is `6 pi x`
B. de-Broglie wavelength is`2pi x`
C. Velocity of electron is `(h)/( 6 pi x m)`
D. Velocity of electron is `(h)/( 2pixm)`

Answer

To solve the problem step by step, we need to analyze the given information about the electron in the hydrogen atom and apply the relevant formulas.

### Step 1: Understand the radius of the orbits
The radius of the nth orbit in a hydrogen atom can be calculated using the formula:
\[ r_n = r_1 \cdot \frac{n^2}{Z} \]
where:
- \( r_1 \) is the radius of the first orbit,
- \( n \) is the principal quantum number (in this case, \( n = 3 \)),
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)).

Given that the radius of the first orbit is \( x \):
\[ r_1 = x \]
Thus, for the third orbit:
\[ r_3 = x \cdot \frac{3^2}{1} = 9x \]

### Step 2: Calculate the de Broglie wavelength
The de Broglie wavelength (\( \lambda \)) of an electron can be calculated using the formula:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the electron,
- \( v \) is the velocity of the electron.

According to Bohr's theory, the angular momentum of the electron is quantized:
\[ mvr = \frac{nh}{2\pi} \]
From this, we can express \( mv \):
\[ mv = \frac{nh}{2\pi r} \]

Substituting this into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{mv} = \frac{h}{\frac{nh}{2\pi r}} = \frac{2\pi r}{n} \]

Now substituting \( r = r_3 = 9x \) and \( n = 3 \):
\[ \lambda = \frac{2\pi (9x)}{3} = 6\pi x \]

### Step 3: Calculate the velocity of the electron
Using the de Broglie wavelength calculated, we can find the velocity:
From the de Broglie wavelength formula:
\[ v = \frac{h}{m\lambda} \]
Substituting \( \lambda = 6\pi x \):
\[ v = \frac{h}{m(6\pi x)} \]

### Final Answers
1. The de Broglie wavelength of the electron in the 3rd orbit is \( 6\pi x \).
2. The velocity of the electron in the 3rd orbit is \( \frac{h}{6\pi mx} \).

### Summary of Correct Options
- De Broglie wavelength: \( 6\pi x \)
- Velocity of the electron: \( \frac{h}{6\pi mx} \)