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Friday, 9 October

JEE Main 2020 · PhysicsMultiple choiceSingle correctMediumMulti-step

JEE Main 8 January 2020, Shift 1, Physics Q17

Question 17 of 75 in this shift, Physics question 17 of 25, Section A.

When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TAT_A eV and de-Broglie wavelength λA\lambda_A. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is TB=(TA−1.5)T_B = (T_A - 1.5) eV. If the de-Broglie wavelength of these photoelectrons λB=2λA\lambda_B = 2\lambda_A, then the work function of metal B is :
  1. (1)1.5 eV
  2. (2)4 eVOfficial answer
  3. (3)3 eV
  4. (4)2 eV

Official answer

Option 2

NTA final key (Jan 2020).