When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy $T_A$ expressed in eV and de-Broglie wavelength $λ_A$. The maximum kinetic energy of photoelectron liberated from another metal B by photons of energy 4.70 eV is $T_B=(T_A-1.50)eV$. If the de- Broglie wavelength of these photoelectrons is $λ_B=2λ_A$, then choose the wrong option:
Answer & explanation
Correct answer: option 4
We know, $K_{max}=E- W$
$∴T_A = 4.25 −W_A$ (i)
$T_B = (T −1.50) = 4.70 −W_B$ (ii)
From these two equations, we have
$W_B −W_A =1.95eV$ (iii)
de – Broglie wavelength is given by
$λ=\frac{h}{\sqrt{2Km}}$ or $λ∝\frac{1}{\sqrt{K}}$
$∴\frac{λ_B}{λ_A}=\sqrt{\frac{K_A}{K_B}}$ $⇒2=\sqrt{\frac{T_A}{T_A-1.5}}$
$⇒T_A=2eV,W_A=2.25eV,W_B=4.20eV$ and $T_B=0.5eV$