Question

When light with a frequency f1 = 547.5 THz illuminates a metal surface, the most energetic photoelectrons have 1.260 x 10^-19 J of kinetic energy. When light with a frequency f2 = 738.8 THz is used instead, the most energetic photo-electrons have 2.480 x 10^-19 J of kinetic energy

Using these experimental results, determine the approximate value of Planck's constant.

Express your answer using four significant figures.

Answer 1

`h=6.377*10^(-34)kgm^2/s`

Step-by-step explanation:

The maximum kinetic energy of the photoelectrons is given by the formula
`K_M=hf-\phi`.

We have two situations where for
`f_1=547.5*10^(12)Hz` we get
`K_(M1)=1.26*10^(-19)J` and for
`f_2=738.8*10^(12)Hz` we get
`K_(M2)=2.48*10^(-19)J`, so we have:

`K_(M1)=hf_1-\phi`

`K_(M2)=hf_2-\phi`

We can eliminate
`\phi` by substracting the first equation to the second:

`K_(M2)-K_(M1)=hf_2-\phi-(hf_1-\phi)=h(f_2-f_1)`

Which means:

`h=(K_(M2)-K_(M1))/(f_2-f_1)=(2.48*10^(-19)J-1.26*10^(-19)J)/(738.8*10^(12)Hz-547.5*10^(12)Hz)=6.377*10^(-34)kgm^2/s`