Photoelectric work- function of a metal is 1 eV. Light of wavelength `lambda = 3000 Å` falls on it. The photoelectrons come out with maximum velocity

Options

A. 10 m/s
B. `10^3m//s
C. `10^4` m//s`
D. `10^6 m//s`

Explanation

To solve the problem step by step, we need to calculate the maximum velocity of photoelectrons emitted from a metal when light of a specific wavelength falls on it. The work function of the metal is given, and we will use the photoelectric effect principles to find the solution.

### Step 1: Understand the relationship between photon energy, work function, and kinetic energy
The energy of a photon (E) can be expressed as:
\[ E = h \nu \]
where \( h \) is Planck's constant and \( \nu \) is the frequency of the light. The frequency can be related to the wavelength (\( \lambda \)) by:
\[ \nu = \frac{c}{\lambda} \]
where \( c \) is the speed of light.

### Step 2: Calculate the energy of the photon
Substituting for \( \nu \):
\[ E = h \frac{c}{\lambda} \]

Given:
- \( h = 6.626 \times 10^{-34} \, \text{J s} \)
- \( c = 3 \times 10^8 \, \text{m/s} \)
- \( \lambda = 3000 \, \text{Å} = 3000 \times 10^{-10} \, \text{m} \)

Now, substituting these values into the equation:
\[ E = 6.626 \times 10^{-34} \times \frac{3 \times 10^8}{3000 \times 10^{-10}} \]

### Step 3: Calculate the numerical value of the photon energy
Calculating the above expression:
\[ E = 6.626 \times 10^{-34} \times \frac{3 \times 10^8}{3 \times 10^{-7}} \]
\[ E = 6.626 \times 10^{-34} \times 10^{15} \]
\[ E = 6.626 \times 10^{-19} \, \text{J} \]

### Step 4: Convert work function from eV to Joules
The work function \( \phi \) is given as 1 eV. To convert this to Joules:
\[ \phi = 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]

### Step 5: Calculate the maximum kinetic energy of the emitted photoelectrons
Using the photoelectric effect equation:
\[ KE_{\text{max}} = E - \phi \]
Substituting the values:
\[ KE_{\text{max}} = 6.626 \times 10^{-19} - 1.6 \times 10^{-19} \]
\[ KE_{\text{max}} = 5.026 \times 10^{-19} \, \text{J} \]

### Step 6: Relate kinetic energy to maximum velocity
The maximum kinetic energy can also be expressed as:
\[ KE_{\text{max}} = \frac{1}{2} mv^2 \]
where \( m \) is the mass of an electron (\( m = 9.1 \times 10^{-31} \, \text{kg} \)). Rearranging gives:
\[ v^2 = \frac{2 KE_{\text{max}}}{m} \]

### Step 7: Substitute values to find maximum velocity
Substituting the values:
\[ v^2 = \frac{2 \times 5.026 \times 10^{-19}}{9.1 \times 10^{-31}} \]
Calculating this gives:
\[ v^2 = \frac{1.0052 \times 10^{-18}}{9.1 \times 10^{-31}} \]
\[ v^2 \approx 1.103 \times 10^{12} \]
Taking the square root:
\[ v \approx 10^6 \, \text{m/s} \]

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