Answer
Gold $(4.82eV)$
Work Step by Step
We are given that
Wavelength of light $\lambda$$=$ $238nm$$=$$238$$\times10$$^-$$^9$m
Speed of the ejected electron
$v=3.75\times10^5$ $m/s$
Energy of the incident Photon
$E=hc/\lambda$
$E=$ $6.626\times10^-$$^3$$^4$ $\times$$3\times10^8$$/$$238\times10^-$$^9$
$E=$ $8.336\times10^-$$^9$$j$
$E=$ $5.21eV$
Kinetic Energy of the ejected electron
$K.E$$=$ $1/2mv^2$
$K.E=$ $1/2$ $\times$ $9.1\times10^-$$^3$$^1$$\times$$(3.75\times10^5$$)$$^2$
$K.E=$$0.624\times10^-$$^1$$^9$$j$
$K.E=$$0.39eV$
As we know
$E=1/2mv^2+\phi$
$\phi$$=$$E-1/2mv^2$
$\phi$$=$ $5.21eV-0.39eV$
$\phi$$=$$4.82$$eV$ which is gold.
$\phi=$ workfunction
$1eV=1.6\times10$$^-$$^1$$^9$$j$
