in an experimental set up to study the photoelectric effect a point soure fo light of power `3.2xx10^(-3)` W was taken. The source can emit monoenergetic photons of energy 5eV and is located at a distance of 0.8 m from the centre of a stationary metallic sphere of work-function 3.0 eV. The radius of the sphere is `r = 8..10^(-3)` m. The efficiency of photoelectric emission is one for every `10^6` incident photons. Based on the information given above answer the questions given below. (Assume that the sphere is isolated and photoelectrons are instantly swepts away after the emission).
de-Broglie wavelength of the fastest moving photoelectron is

Options

A. `6.63 Å`
B. `8.69 Å`
C. `2 Å`
D. `5.26 Å`

Explanation

To find the de-Broglie wavelength of the fastest moving photoelectron, we will follow these steps:

### Step 1: Calculate the energy of the emitted photons
The energy of the emitted photons is given as \( E = 5 \, \text{eV} \).

### Step 2: Calculate the maximum kinetic energy of the emitted photoelectrons
The maximum kinetic energy (\( K_{\text{max}} \)) of the photoelectrons can be calculated using the formula:
\[ K_{\text{max}} = E - \phi \]
where \( \phi \) is the work function of the metallic sphere. Given that \( \phi = 3 \, \text{eV} \):
\[ K_{\text{max}} = 5 \, \text{eV} - 3 \, \text{eV} = 2 \, \text{eV} \]

### Step 3: Convert kinetic energy from eV to Joules
To convert \( K_{\text{max}} \) from electron volts to Joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ K_{\text{max}} = 2 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 3.2 \times 10^{-19} \, \text{J} \]

### Step 4: Use the de-Broglie wavelength formula
The de-Broglie wavelength \( \lambda \) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum of the electron. The momentum can be expressed in terms of kinetic energy as:
\[ p = \sqrt{2m_e K_{\text{max}}} \]
Substituting this into the de-Broglie wavelength formula gives:
\[ \lambda = \frac{h}{\sqrt{2m_e K_{\text{max}}}} \]

### Step 5: Substitute known values
- Planck's constant \( h = 6.63 \times 10^{-34} \, \text{Js} \)
- Mass of the electron \( m_e = 9.1 \times 10^{-31} \, \text{kg} \)
- \( K_{\text{max}} = 3.2 \times 10^{-19} \, \text{J} \)

Now we can substitute these values into the equation:
\[ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 3.2 \times 10^{-19}}} \]

### Step 6: Calculate the denominator
Calculating the denominator:
\[ 2 \times 9.1 \times 10^{-31} \times 3.2 \times 10^{-19} = 5.824 \times 10^{-48} \]
Taking the square root:
\[ \sqrt{5.824 \times 10^{-48}} \approx 7.62 \times 10^{-24} \]

### Step 7: Calculate the wavelength
Now substituting back into the wavelength equation:
\[ \lambda = \frac{6.63 \times 10^{-34}}{7.62 \times 10^{-24}} \approx 8.69 \times 10^{-10} \, \text{m} \]
Converting to Angstroms (1 Angstrom = \( 10^{-10} \, \text{m} \)):
\[ \lambda \approx 8.69 \, \text{Å} \]

### Final Answer
The de-Broglie wavelength of the fastest moving photoelectron is approximately \( 8.69 \, \text{Å} \).

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