Light of wavelength `2000 Å` falls on an aluminium surface . In aluminium `4.2 e V` of energy is required to remove an electron from its surface. What is the kinetic energy , in electron volt of (a) the fastest and (b) the slowest emitted photo-electron . ( c) What is the stopping potential ? (d) What is the cut - off wavelength for aluminum? (Plank's constant `h = 6.6 xx 10^(-34) J-s` and speed of light `c = 3 xx 10^(8) m s^(-1).

Explanation

To solve the problem step by step, we will follow the instructions given in the video transcript and apply the relevant physics concepts.

### Given Data:
- Wavelength of light, \( \lambda = 2000 \, \text{Å} = 2000 \times 10^{-10} \, \text{m} \)
- Work function of aluminum, \( \Phi = 4.2 \, \text{eV} \)
- Planck's constant, \( h = 6.63 \times 10^{-34} \, \text{J s} \)
- Speed of light, \( c = 3 \times 10^{8} \, \text{m/s} \)
- Conversion factor: \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \)
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