When photon of energy 4.0 eV strikes the surface of a metal A , the ejected photoelectrons have maximum kinetic energy $\mathrm{T}_{\mathrm{A}} \mathrm{eV}$ and de-Broglie wavelength $\lambda_{\mathrm{A}}$.The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is $\mathrm{T}_{\mathrm{B}}=\left(\mathrm{T}_{\mathrm{A}}-1.5\right) \mathrm{eV}$. If the de-Broglie waelength of these photoelectrons $\lambda_{\mathrm{B}}=2 \lambda_{\mathrm{A}}$, then the work function of metal B is :