Dual Nature of Matter

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68 of 200

Question 68 of 200

When photon of energy 4.0eV4.0eV strikes the surface of a metal A,A, the ejected photoelectrons have maximum kinetic energy TAeV{T}_{A}eV and de-Broglie wavelength λA.{\lambda }_{A}. The maximum kinetic energy of photoelectrons liberated from another metal BB by photon of energy 4.50eV4.50eV is TB=(TA−1.5)eV.{T}_{B}=\left({T}_{A}-1.5\right)eV. If the de-Broglie wavelength of these photoelectrons λB=2λA,{\lambda }_{B}=2{\lambda }_{A}, then the work function of metal BB is:

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Answer: (A) 4eV4eV

Relation between de-Broglie wavelength and K.E. is
λ=h2(KE)me⇒λ∝1KE\lambda =\dfrac{h}{\sqrt{2(KE){m}_{e}}}\Rightarrow \lambda \propto \dfrac{1}{\sqrt{KE}}
λAλB=KEBKEA\dfrac{{\lambda }_{A}}{{\lambda }_{B}}=\dfrac{\sqrt{K{E}_{B}}}{\sqrt{K{E}_{A}}}
⇒12=TA−1.5TA⇒TA=2eV\Rightarrow \dfrac{1}{2}=\sqrt{\dfrac{{T}_{A}-1.5}{{T}_{A}}}\Rightarrow {T}_{A}=2eV
∴KEB=2−1.5=0.5eV∴K{E}_{B}=2-1.5=0.5eV
ϕB=4.5−0.5=4eV{\phi }_{B}=4.5-0.5=4eV

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