Question

Red light of wavelength 651 nm produces photoelectrons from a certain photoemissive material. Green light of wavelength 521 nm produces photoelectrons from the same material with 1.50 times the maximum kinetic energy. What is the material's work function?

Answer 1

material work function is 0.956 eV

Step-by-step explanation:

given data

red wavelength 651 nm

green wavelength 521 nm

photo electrons = 1.50 × maximum kinetic energy

to find out

material work function

solution

we know by Einstein photo electric equation that is

for red light

h ( c / λr ) = Ф + kinetic energy

for green light

h ( c / λg ) = Ф + 1.50 × kinetic energy

now from both equation put kinetic energy from red to green

h ( c / λg ) = Ф + 1.50 × (h ( c / λr ) - Ф)

Ф =( hc / 0.50) × ( 1.50/ λr - 1/ λg)

put all value

Ф =( 6.63 ×
`10^(-34)` (3 ×
`10^(8)` ) / 0.50) × ( 1.50/ λr - 1/ λg)

Ф =( 6.63 ×
`10^(-34)` (3 ×
`10^(8)` ) / 0.50 ) × ( 1.50/ 651×
`10^(-9)` - 1/ 521 ×
`10^(-9)`)

Ф = 1.5305 ×
`10^(-19)` J × ( 1ev / 1.6 ×
`10^(-19)` J )

Ф = 0.956 eV

material work function is 0.956 eV

Answer 2

Given:

wavelength of red light,
`\lambda _(R) = 651 nm`

wavelength of red light,
`\lambda _(G) = 521 nm`

Kinetic energy of green light,
`K.E _(G) = 1.5K.E_(max)`

Solution:

To calculate the work function of the material,
`\phi`

of the material can be calculated by using Einstein's equation for photoelectric effect:

`K.E = (hc)/(\lambda ) - \phi` (1)

where,

h = Plank's constant =
`6.64* 10^(-34) J-s`

`\lambda` = wavelength of light

`\phi` = work function of material

c = speed of light in vacuum =
`3* 10^(8) m/s`

Now, for red light:

`K.E_(max) = (hc)/(\lambda_(R) ) - \phi` (2)

Now, for green light:

`1.5K.E_(max) = (hc)/(\lambda_(G)) - \phi`

`K.E_(max) = (2)/(3)(hc)/(\lambda_(G) ) - \phi` (3)

Using eqn (2) and (3):

`(1)/(3)\phi = hc((1)/(\lambda R) - (2)/(3\lambda_(G)))`

`\phi = 6.64* 10^(-34)* 3* 10^(8)((1)/(651* 10^(-9)) - (2)/(3* 521^(-9)))`

`\phi = 1.532* 10^(-19) J`

`\phi = (1.532* 10^(-19))/(1.6* 10^(-19))`

`\phi = 0.958 eV`

Answer:

Work function of the material:

`\phi = 1.532* 10^(-19) J` or 0.958 eV