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

Maximum wavelength will be
`3.96* 10^(-7)m`

Step-by-step explanation:

It is given wavelength
`\lambda =242nm=242* 10^(-9)m`

Speed of light
`c=3* 10^8m/sec`

Plank's constant
`h=6.6* 10^(-4)Js`

So energy is equal to

`E=(hc)/(\lambda )=(6.6* 10^(-34)* 3* 10^8)/(242* 10^(-9))=8.18* 10^(-19)J`

Maximum kinetic energy is given

`KE_(max)=1.99eV=1.99* 1.6* 10^(-19)=3.184* 10^(-19)J`

Work function is equal to

`w_0=E-KE_(max)=8.18* 10^(-19)-3.184* 10^(-19)=5* 10^(-19)J`

`(hc)/(\lambda _0)=5* 10^(-19)`

`(6.6* 10^(-34)* 3* 10^8)/(\lambda _0)=5* 10^(-19)`

`\lambda _0=3.96* 10^(-7)m`