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Question

A mercury arc lamp provides 0.1 W of ultra-violet radiation at a wavelength of  λ=2537A only. The photo tube (cathode of photo electric device) consists of potassium and has an effective area of 4 cm2. The cathode is located at a distance of 1m from the radiation source. The work function for potassium is ϕ0=2.2 eV.
What is the cut off potential V0 ?
  1. 2.69 V
  2. 1.35 V
  3. 26.9 V
  4. 5.33 V


Solution

The correct option is A 2.69 V
Cut-off potential V0 found by photoelectric equation i.e.
Epϕ=eV0
We have Ep as
Ep=hcλ=12.43×1072537×10104.9 eV
ϕ=2.2 eV
Putting these values in photoelectric equation we get
2.7 eV=eV0
V0=2.7 V

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