Question

When light of wavelength 242 nm shines on a metal surface the maximum kinetic energy of the photoelectrons is 1.99 eV. What is the maximum wavelength (in nm) of light that will produce photoelectrons from this surface? (Use 1 eV = 1.602 ✕ 10^−19 J, e = 1.602 ✕ 10^−19 C, c = 2.998 ✕ 10^8 m/s, and h = 6.626 ✕ 10^−34 J · s = 4.136 ✕ 10^−15 eV · s as necessary.)

Answer 1

Answer:

Maximum wavelength will be $3.96\times 10^{-7}m$

Explanation:

It is given wavelength $\lambda =242nm=242\times 10^{-9}m$

Speed of light $c=3\times 10^8m/sec$

Plank's constant $h=6.6\times 10^{-4}Js$

So energy is equal to

$E=\frac{hc}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{242\times 10^{-9}}=8.18\times 10^{-19}J$

Maximum kinetic energy is given

$KE_{max}=1.99eV=1.99\times 1.6\times 10^{-19}=3.184\times 10^{-19}J$

Work function is equal to

$w_0=E-KE_{max}=8.18\times 10^{-19}-3.184\times 10^{-19}=5\times 10^{-19}J$

$\frac{hc}{\lambda _0}=5\times 10^{-19}$

$\frac{6.6\times 10^{-34}\times 3\times 10^8}{\lambda _0}=5\times 10^{-19}$

$\lambda _0=3.96\times 10^{-7}m$