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 find the maximum velocity of photoelectrons emitted from a metal with a work function of 1 eV when illuminated by light of wavelength 3000 Å, we can follow these steps:

### Step 1: Calculate the Energy of the Incident Light
The energy of the incident light can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \(h\) (Planck's constant) = \(6.626 \times 10^{-34} \, \text{Js}\)
- \(c\) (speed of light) = \(3.00 \times 10^8 \, \text{m/s}\)
- \(\lambda\) (wavelength) = \(3000 \, \text{Å} = 3000 \times 10^{-10} \, \text{m}\)

Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3.00 \times 10^8 \, \text{m/s})}{3000 \times 10^{-10} \, \text{m}} \]

### Step 2: Convert Wavelength to Energy in eV
Alternatively, we can use the formula for energy in electron volts when the wavelength is given in angstroms:
\[ E = \frac{12400}{\lambda} \, \text{eV} \]
Substituting \(\lambda = 3000 \, \text{Å}\):
\[ E = \frac{12400}{3000} \approx 4.13 \, \text{eV} \]

### Step 3: Calculate the Maximum Kinetic Energy of the Photoelectrons
The maximum kinetic energy (K.E.) of the emitted photoelectrons can be calculated using the photoelectric equation:
\[ \text{K.E.} = E - \phi \]
where \(\phi\) is the work function of the metal. Given that \(\phi = 1 \, \text{eV}\):
\[ \text{K.E.} = 4.13 \, \text{eV} - 1 \, \text{eV} = 3.13 \, \text{eV} \]

### Step 4: Convert Kinetic Energy to Joules
To convert kinetic energy from electron volts to joules, use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):
\[ \text{K.E.} = 3.13 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 5.008 \times 10^{-19} \, \text{J} \]

### Step 5: Calculate the Maximum Velocity
Using the formula for kinetic energy:
\[ \text{K.E.} = \frac{1}{2} mv^2 \]
we can solve for \(v\):
\[ v = \sqrt{\frac{2 \times \text{K.E.}}{m}} \]
where \(m\) (mass of an electron) = \(9.11 \times 10^{-31} \, \text{kg}\):
\[ v = \sqrt{\frac{2 \times 5.008 \times 10^{-19} \, \text{J}}{9.11 \times 10^{-31} \, \text{kg}}} \]

### Step 6: Perform the Calculation
Calculating the above expression:
\[ v = \sqrt{\frac{1.0016 \times 10^{-18}}{9.11 \times 10^{-31}}} \approx \sqrt{1.099 \times 10^{12}} \approx 1.05 \times 10^6 \, \text{m/s} \]

### Final Answer
The maximum velocity of the photoelectrons is approximately \(1.05 \times 10^6 \, \text{m/s}\).

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