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

Answer:

$h=6.377\times10^{-34}kgm^2/s$

Explanation:

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

We have two situations where for $f_1=547.5\times10^{12}Hz$ we get $K_{M1}=1.26\times10^{-19}J$ and for $f_2=738.8\times10^{12}Hz$ we get $K_{M2}=2.48\times10^{-19}J$, 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:

$h=\frac{K_{M2}-K_{M1}}{f_2-f_1}=\frac{2.48\times10^{-19}J-1.26\times10^{-19}J}{738.8\times10^{12}Hz-547.5\times10^{12}Hz}=6.377\times10^{-34}kgm^2/s$