In a photoelectric effect experiment you illuminate a metal with light of an unknown wavelength and measure the maximum kinetic energy of the photoelectrons to be 0.65 ev. then you illuminate the same metal with light of a wavelength known to be 2/3 of the first wavelength and measure a maximum kinetic energy of 2.3 ev for the photoelectrons.

(a) find the first wavelength, in nanometers.
(b) find the metal's work function, in electron volts.

(a). The first wavelength is 401.0 nm.

(b). The metal's work function is 2.55 eV.

Explanation:

Given that,

Maximum kinetic energy = 0.65 eV

Second wavelength

(a). We need to calculate the wavelength

Using equation of work function for first wavelength

.....(I)

For second wavelength,

Put the value of second wavelength

....(II)

By subtraction equation (I) from (II)

(b). We need to calculate the work function

Using formula of work function

Put the value into the formula

Hence, (a). The first wavelength is 401.0 nm.

(b). The metal's work function is 2.55 eV.

increase masses, decrease separation

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.