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Light of wavelength 4000Å is allowed to ...

Light of wavelength `4000Å` is allowed to fall on a metal surface having work function 2 eV. The maximum velocity of the emitted electrons is
`(h=6.6xx10^(-34)Js)`

A

`1.35xx10^(5)ms^(-1)`

B

`2.7xx10^(5)ms^(-1)`

C

`6.2xx10^(5)ms^(-1)`

D

`8.1xx10^(5)ms^(-1)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will use the Einstein equation of the photoelectric effect, which relates the energy of the incoming photons to the work function of the metal and the kinetic energy of the emitted electrons. ### Step-by-Step Solution: 1. **Convert Wavelength to Nanometers:** The given wavelength is \( 4000 \, \text{Å} \). \[ 4000 \, \text{Å} = 4000 \times 10^{-10} \, \text{m} = 400 \, \text{nm} \] 2. **Calculate the Energy of the Photon:** Using the formula for the energy of a photon: \[ E = \frac{hc}{\lambda} \] where \( h = 6.63 \times 10^{-34} \, \text{Js} \) and \( c = 3 \times 10^8 \, \text{m/s} \). Substituting the values: \[ E = \frac{(6.63 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{400 \times 10^{-9} \, \text{m}} \] \[ E = \frac{1.989 \times 10^{-25}}{400 \times 10^{-9}} = 4.9725 \times 10^{-19} \, \text{J} \] 3. **Convert Energy to Electron Volts:** To convert joules to electron volts, use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \): \[ E = \frac{4.9725 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.11 \, \text{eV} \] 4. **Calculate the Maximum Kinetic Energy of the Electrons:** The maximum kinetic energy \( KE_{\text{max}} \) of the emitted electrons is given by: \[ KE_{\text{max}} = E - \phi \] where \( \phi \) is the work function of the metal (given as \( 2 \, \text{eV} \)): \[ KE_{\text{max}} = 3.11 \, \text{eV} - 2 \, \text{eV} = 1.11 \, \text{eV} \] 5. **Convert Kinetic Energy to Joules:** \[ KE_{\text{max}} = 1.11 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.776 \times 10^{-19} \, \text{J} \] 6. **Use the Kinetic Energy Formula to Find Velocity:** The kinetic energy is also given by: \[ KE = \frac{1}{2} mv^2 \] Rearranging for \( v \): \[ v = \sqrt{\frac{2 \times KE}{m}} \] where \( m \) is the mass of an electron \( (9.1 \times 10^{-31} \, \text{kg}) \): \[ v = \sqrt{\frac{2 \times 1.776 \times 10^{-19}}{9.1 \times 10^{-31}}} \] \[ v = \sqrt{\frac{3.552 \times 10^{-19}}{9.1 \times 10^{-31}}} = \sqrt{3.90 \times 10^{11}} \approx 6.24 \times 10^5 \, \text{m/s} \] ### Final Answer: The maximum velocity of the emitted electrons is approximately: \[ v \approx 6.24 \times 10^5 \, \text{m/s} \]
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