When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA (expressed in eV ) and de-Broglie wavelength λ. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB=(TA-1.50 eV ). If the de-Broglie wavelength (in eV ) of these photoelectrons is λB=2 λA, then find TB (in .eV ).

Question

When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA (expressed in eV ) and de-Broglie wavelength λ. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is TB=(TA-1.50 eV ). If the de-Broglie wavelength (in eV ) of these photoelectrons is λB=2 λA, then find TB (in .eV ). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

4.25=TA+φA ; 4.70=TB+φB TB=TA-1.5 (Given) λ=(h/√2 m T) and λA=(1/2) λB ⇒ TA=4 TB

4.25=TA​+ϕA​;4.70=TB​+ϕB​ TB​=TA​−1.5 (Given) λ=2mT​h​ and λA​=21​λB​ ⇒TA​=4TB​

$4.25 = T_{A} + ϕ_{A}; 4.70 = T_{B} + ϕ_{B}$
$T_{B} = T_{A} - 1.5$ (Given)
$λ = \frac{h}{2 m T}$
and $λ_{A} = \frac{1}{2} λ_{B}$
$\Rightarrow T_{A} = 4 T_{B}$