Electronmagnetic radiations having `lambda=310Å`are subjected to a metal sheet having work function `=12.8eV`. What will be the velocity of photoelectrons with maximum Kinetic Energy….

Options

A. (a) 0, no emission will occur
B. (b) `4.352xx10^(6)` m/s
C. (c) `3.09xx10^(6)` m/s
D. (d) `8.72xx10^(6)` m/s

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Convert the wavelength from Angstroms to meters
Given:
- Wavelength, \( \lambda = 310 \, \text{Å} = 310 \times 10^{-10} \, \text{m} \)

### Step 2: Calculate the energy of the incoming photons
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) (Planck's constant) = \( 6.626 \times 10^{-34} \, \text{Js} \)
- \( c \) (speed of light) = \( 3 \times 10^8 \, \text{m/s} \)

Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{310 \times 10^{-10} \, \text{m}} \]

Calculating this gives:
\[ E \approx 6.4 \times 10^{-19} \, \text{J} \]

### Step 3: Convert energy from Joules to electron volts
To convert Joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E \approx \frac{6.4 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 40 \, \text{eV} \]

### Step 4: Calculate the maximum kinetic energy of the photoelectrons
The maximum kinetic energy (KE) of the photoelectrons is given by:
\[ KE = E - \phi \]
Where \( \phi \) is the work function of the metal.

Given:
- Work function, \( \phi = 12.8 \, \text{eV} \)

Substituting the values:
\[ KE = 40 \, \text{eV} - 12.8 \, \text{eV} = 27.2 \, \text{eV} \]

### Step 5: Convert kinetic energy from electron volts to Joules
Using the conversion factor:
\[ KE \approx 27.2 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} \approx 4.352 \times 10^{-18} \, \text{J} \]

### Step 6: Use the kinetic energy to find the velocity of the photoelectrons
The kinetic energy is also given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
Where \( m \) is the mass of the electron (\( m = 9.1 \times 10^{-31} \, \text{kg} \)).

Rearranging for \( v \):
\[ v = \sqrt{\frac{2 \times KE}{m}} \]

Substituting the values:
\[ v = \sqrt{\frac{2 \times 4.352 \times 10^{-18} \, \text{J}}{9.1 \times 10^{-31} \, \text{kg}}} \]

Calculating this gives:
\[ v \approx 3.09 \times 10^6 \, \text{m/s} \]

### Final Answer
The velocity of the photoelectrons with maximum kinetic energy is approximately \( 3.09 \times 10^6 \, \text{m/s} \).

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