de Broglie proposed dual nature for electron by putting his famous equation `lambda = (h)/(mv).` Later on, Heisenberg proposed uncertainty principle as `deltapDeltax ge (h)/(4pi).` On the contrary,
Particle nature of electron was established on the basis of photoelectric effect. When a photon strikes the metal surface it gives up its energy to the electron. Part of this energy (say W) is used by the electrons to escape from the metal and the remaining energy imparts kinetic energy `(1//2 mv^(2))` to the ejected photoelectron. The potential applied on the surface to reduce the velocity of photoelectron to zero is known as stopping potential .
The circumference of third orbit of a single electron species is 3 nm. What may be the approximate wavelength of the photon required to just ionize electron from this orbit?

Options

A. `91.1 nm`
B. `364.7 nm`
C. `821 nm`
D. `205 nm`

Explanation

To solve the problem, we need to find the approximate wavelength of the photon required to just ionize an electron from the third orbit of a single electron species, given that the circumference of this orbit is 3 nm.

### Step-by-Step Solution:

1. **Identify the Circumference**:
The circumference of the third orbit is given as:
\[ C = 3 \text{ nm} = 3 \times 10^{-9} \text{ m} \]

2. **Relate Circumference to Radius**:
The circumference of a circular orbit is given by the formula:
\[ C = 2\pi r \]
Rearranging this to find the radius \( r \):
\[ r = \frac{C}{2\pi} = \frac{3 \times 10^{-9}}{2\pi} \]
Calculating this gives:
\[ r \approx \frac{3 \times 10^{-9}}{6.2832} \approx 4.77 \times 10^{-10} \text{ m} = 4.77 \text{ Å} \]

3. **Use the Radius to Find Z**:
The radius of the orbit can also be expressed using the formula:
\[ r = \frac{0.529 \times n^2}{Z} \]
For the third orbit, \( n = 3 \):
\[ 4.77 = \frac{0.529 \times 3^2}{Z} \]
Simplifying this:
\[ 4.77 = \frac{0.529 \times 9}{Z} \]
\[ Z = \frac{0.529 \times 9}{4.77} \]
Calculating \( Z \):
\[ Z \approx 1 \]

4. **Calculate the Energy of the Electron**:
The energy of the electron in the nth orbit is given by:
\[ E = -\frac{13.6 Z^2}{n^2} \]
Substituting \( Z = 1 \) and \( n = 3 \):
\[ E = -\frac{13.6 \times 1^2}{3^2} = -\frac{13.6}{9} \approx -1.51 \text{ eV} \]

5. **Calculate the Wavelength**:
The energy of the photon required to ionize the electron is equal to the absolute value of the energy of the electron:
\[ E = 1.51 \text{ eV} \]
The wavelength \( \lambda \) of the photon can be calculated using the formula:
\[ \lambda = \frac{1240}{E} \text{ (in nm)} \]
Substituting \( E = 1.51 \):
\[ \lambda = \frac{1240}{1.51} \approx 821 \text{ nm} \]

### Final Answer:
The approximate wavelength of the photon required to just ionize the electron from this orbit is:
\[ \lambda \approx 821 \text{ nm} \]