In 1924, de-Broglie proposed that every particle possesses wave properties with a wavelength , `lambda` given by `lambda = h/(mv)` where m is the mass of the particle, `v` is its velocity and h is Planck's constant. The de-Broglie prediction was confirmed experimentally when it was found that an electron beam undergoes diffraction , a phenomenon characteristic of waves. The de-Broglie wavelength can be estimated by measuring kinetic energy of an electron accelerating by a potential V as :
`1/2 mv^2 = eV "where" 1eV = 1.6 xx 10^(-19) J, h = 6.6 xx 10^(-34) Js`.
If `lambda` is the wavelength associated with the electron in the 4th circular orbit of hydrogen atom, then radius of the orbit is

Options

A. `lambda/(2pi)`
B. `2(lambda)/(pi)`
C. `2/(pi lambda)`
D. `(2 pi)/(lambda)`

Explanation

To find the radius of the 4th circular orbit of a hydrogen atom using the de-Broglie wavelength, we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and potential difference
The kinetic energy (KE) of an electron accelerating through a potential difference \( V \) is given by:
\[ \frac{1}{2} mv^2 = eV \]
where \( e \) is the charge of the electron (1 eV = \( 1.6 \times 10^{-19} \) J).

### Step 2: Express velocity in terms of potential difference
From the kinetic energy equation, we can solve for \( v \):
\[ v = \sqrt{\frac{2eV}{m}} \]

### Step 3: Use de-Broglie's wavelength formula
The de-Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{mv} \]
Substituting the expression for \( v \) from Step 2:
\[ \lambda = \frac{h}{m \sqrt{\frac{2eV}{m}}} = \frac{h \sqrt{m}}{\sqrt{2eV}} \]

### Step 4: Relate the radius of the orbit to angular momentum
For a circular orbit, the angular momentum \( L \) is quantized and given by:
\[ L = mvr = n\frac{h}{2\pi} \]
For the 4th orbit, \( n = 4 \):
\[ mvr = 4\frac{h}{2\pi} = \frac{2h}{\pi} \]

### Step 5: Solve for the radius \( r \)
Now we can express \( r \) in terms of \( \lambda \):
\[ r = \frac{2h}{\pi mv} \]
Substituting \( mv \) from the de-Broglie wavelength:
\[ mv = \frac{h}{\lambda} \]
Thus,
\[ r = \frac{2h}{\pi \left(\frac{h}{\lambda}\right)} = \frac{2\lambda}{\pi} \]

### Conclusion
The radius of the 4th circular orbit of the hydrogen atom is:
\[ r = \frac{2\lambda}{\pi} \]