To solve the problem, we need to calculate the maximum kinetic energy (K.E.) of the ejected electrons when light of a specific wavelength falls on a metal surface with a known work function. Here's a step-by-step solution:
### Step 1: Understand the Concept
The photoelectric effect states that when light falls on a metal surface, electrons can be ejected if the energy of the incident light is greater than the work function (φ) of the metal. The maximum kinetic energy of the ejected electrons can be calculated using the photoelectric equation:
\[ E = K.E. + \phi \]
where:
- \( E \) is the energy of the incident light,
- \( K.E. \) is the maximum kinetic energy of the ejected electrons,
- \( \phi \) is the work function of the metal.
### Step 2: Calculate the Energy of the Incident Light
The energy of the incident light can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength of the light in meters.
Given that the wavelength \( \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} \).
### Step 3: Substitute the Values
Now, substituting the values into the equation:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{5000 \times 10^{-10} \, \text{m}} \]
### Step 4: Calculate the Energy
Calculating the above expression:
\[ E = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{5000 \times 10^{-10}} \]
\[ E = \frac{1.9878 \times 10^{-25}}{5 \times 10^{-7}} \]
\[ E = 3.9756 \times 10^{-19} \, \text{J} \]
### Step 5: Convert Energy to Electron Volts
To convert Joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E = \frac{3.9756 \times 10^{-19}}{1.6 \times 10^{-19}} \]
\[ E \approx 2.485 \, \text{eV} \]
### Step 6: Apply the Photoelectric Equation
Now, using the photoelectric equation:
\[ E = K.E. + \phi \]
We know \( \phi = 1.5 \, \text{eV} \):
\[ 2.485 \, \text{eV} = K.E. + 1.5 \, \text{eV} \]
### Step 7: Solve for Maximum K.E.
Rearranging gives:
\[ K.E. = 2.485 \, \text{eV} - 1.5 \, \text{eV} \]
\[ K.E. = 0.985 \, \text{eV} \]
### Final Answer
The maximum kinetic energy of the ejected electrons is approximately \( 0.985 \, \text{eV} \).
