Question

The work function for a metal surface is 4.98 eV. What is the largest wavelength of light in nm that will produce photoelectrons from this surface? (Use 1 eV = 1.602 ✕ 10−19 J, e = 1.602 ✕ 10−19 C, c = 2.998 ✕ 108 m/s, and h = 6.626 ✕ 10−34 J · s = 4.136 ✕ 10−15 eV · s as necessary.)

Answer 1

Answer:$\lambda =248.99 nm$

Explanation:

Given

Work function$\left ( \phi \right )=4.98\approx 1.602\times 10^{-19}\times 4.98$

$h=6.626\times 10^{-34} J$

$c=2.998\times 10^8$

$\phi =\frac{hc}{\lambda }$

$\lambda =\frac{hc}{\phi }$

$\lambda =\frac{6.626\times 10^{-34}\times 2.998\times 10^8}{4.98\times 1.602\times 10^{-19}}$

$\lambda =248.99 nm$