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

Answer:

material work function is 0.956 eV

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:

$K.E = \frac{hc}{\lambda } - \phi$                                    (1)

where,

h = Plank's constant = $6.64\times 10^{-34} J-s$

$\lambda$ = wavelength of light

$\phi$ = work function of material

c = speed of light in vacuum =  $3\times 10^{8} m/s$

Now, for red light:

$K.E_{max} = \frac{hc}{\lambda_{R} } - \phi$                   (2)

Now, for green light:

$1.5K.E_{max} = \frac{hc}{\lambda_{G}} - \phi$

$K.E_{max} = \frac{2}{3}\frac{hc}{\lambda_{G} } - \phi$     (3)

Using eqn (2) and (3):

$\frac{1}{3}\phi = hc(\frac{1}{\lambda R} - \frac{2}{3\lambda_{G}})$

$\phi = 6.64\times 10^{-34}\times 3\times 10^{8}(\frac{1}{651\times 10^{-9}} - \frac{2}{3\times 521^{-9}})$

$\phi = 1.532\times 10^{-19} J$

$\phi = \frac{1.532\times 10^{-19}}{1.6\times 10^{-19}}$

$\phi = 0.958 eV$

Answer:

Work function of the material:

$\phi = 1.532\times 10^{-19} J$  or 0.958 eV