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Question

When photons of energy 4.25 eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy, TA expressed in eV and de-Broglie wavelength λA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB=(TA1.50 eV). If the de-Broglie wavelength of these photoelectrons is λB=2λA, then
  1. The work function of A is 2.25 eV
  2. The work function of B is 4.20 eV
  3. TA=2.0 eV
  4. TB=2.75 eV


Solution

The correct options are
A The work function of A is 2.25 eV
B The work function of B is 4.20 eV
C TA=2.0 eV
From photoelectric equation for the two metals
4.25 eVϕA=TA=p2A2m=h22mλ2A
and 4.70 eVϕB=TB=h22mλ2B
TATB=(λBλA)2=4, given TB=TA1.50 eV
On solving we get, TA=2.0 eV and TB=0.50 eV
ϕA=2.25 eV,ϕB=4.20 eV

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