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 .
When a beam of photons of a particular energy was incident on a surface of a particular pure metal having work function =(40 eV), some emitted photoelectrons had stopping potential equal to 22 V. some had 12V and rest had lower values. Calculate the wavelength of incident photons assuming that at least one photoelectron is ejected with maximum possible kinetic energy :

Options

A. `310 "Å"`
B. `298 "Å"`
C. `238 "Å"`
D. `200 "Å"`

Answer

To solve the problem, we need to calculate the wavelength of the incident photons based on the given information about the work function and stopping potential of photoelectrons.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Work function (Φ) of the metal = 40 eV
- Stopping potential (V) for the maximum kinetic energy = 22 V

2. **Calculate the Maximum Kinetic Energy (KE) of the Ejected Photoelectron:**
The maximum kinetic energy of the photoelectron can be calculated using the formula:
\[ KE = e \cdot V \]
where \( e \) is the charge of the electron (approximately 1 eV/V).
\[ KE = 22 \text{ eV} \]

3. **Calculate the Total Energy of the Incident Photon (E):**
The total energy of the incident photon must overcome the work function and provide kinetic energy to the ejected electron. Therefore, we can express this as:
\[ E = \Phi + KE \]
Substituting the known values:
\[ E = 40 \text{ eV} + 22 \text{ eV} = 62 \text{ eV} \]

4. **Convert the Energy of the Photon to Wavelength (λ):**
The energy of a photon is related to its wavelength by the equation:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant (\( 4.135667696 \times 10^{-15} \) eV·s) and \( c \) is the speed of light (\( 3 \times 10^8 \) m/s). Rearranging the formula gives:
\[ \lambda = \frac{hc}{E} \]
First, we convert \( E \) from eV to Joules (1 eV = \( 1.602 \times 10^{-19} \) J):
\[ E = 62 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} = 9.93 \times 10^{-18} \text{ J} \]
Now substituting the values into the wavelength formula:
\[ \lambda = \frac{(4.135667696 \times 10^{-15} \text{ eV·s})(3 \times 10^8 \text{ m/s})}{62 \text{ eV}} \]
\[ \lambda = \frac{1.241 \times 10^{-6} \text{ eV·m}}{62 \text{ eV}} = 2.00 \times 10^{-8} \text{ m} = 200 \text{ nm} \]

5. **Final Result:**
The wavelength of the incident photons is approximately:
\[ \lambda = 200 \text{ nm} \]