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If light of wavelength 6200Å falls on a ...

If light of wavelength 6200Å falls on a photosensitive surface of work function 2 eV, the kinetic energy of the most energetic photoelectron will be

A

0.5 eV

B

1 eV

C

zero

D

0.75 eV

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To solve the problem of finding the kinetic energy of the most energetic photoelectron when light of wavelength 6200 Å falls on a photosensitive surface with a work function of 2 eV, we can follow these steps: ### Step 1: Convert the wavelength from Angstroms to meters The wavelength is given as 6200 Å. We need to convert this to meters for further calculations: \[ \text{Wavelength} (\lambda) = 6200 \, \text{Å} = 6200 \times 10^{-10} \, \text{m} \] ### Step 2: Calculate the energy of the incident 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: \[ E = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{6200 \times 10^{-10} \, \text{m}} \] ### Step 3: Calculate the energy in electron volts To convert the energy from joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \): \[ E \, (\text{in eV}) = \frac{E \, (\text{in J})}{1.602 \times 10^{-19}} \] ### Step 4: Calculate the kinetic energy of the photoelectron The kinetic energy (KE) of the most energetic photoelectron can be calculated using the photoelectric equation: \[ KE = E - W \] where \( W \) is the work function. Given that \( W = 2 \, \text{eV} \): \[ KE = E - 2 \, \text{eV} \] ### Step 5: Substitute the values and calculate After calculating \( E \) in eV from the previous steps, we can substitute it into the kinetic energy equation to find the final result. ### Final Calculation 1. Calculate \( E \) using the values provided. 2. Substitute \( E \) into the kinetic energy equation to find \( KE \). ### Conclusion If the calculated energy \( E \) is equal to the work function \( W \), then: \[ KE = E - W = 0 \, \text{eV} \]

To solve the problem of finding the kinetic energy of the most energetic photoelectron when light of wavelength 6200 Å falls on a photosensitive surface with a work function of 2 eV, we can follow these steps: ### Step 1: Convert the wavelength from Angstroms to meters The wavelength is given as 6200 Å. We need to convert this to meters for further calculations: \[ \text{Wavelength} (\lambda) = 6200 \, \text{Å} = 6200 \times 10^{-10} \, \text{m} \] ...
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MARVEL PUBLICATION-ELECTRONS AND PHOTONS -TEST YOUR GRASP - 17
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