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.

Answer 1

The approximate value of Planck's constant is
`6.377*10^(-34)\ J`

Step-by-step explanation:

Given that,

Frequency
`f_(1)= 547.5\ THz`

Kinetic energy
`K.E=1.260*10^(-19)\ J`

Frequency
`f_(2)=738.8\ THz`

Kinetic energy
`K.E=2.480*10^(-19)\ J`

We need to calculate the approximate value of Planck's constant

Using formula of change in energy

`E = hf`

`K.E_(2)-K.E_(1)=h(f_(2)-f_(1))`

`h=(K.E_(2)-K.E_(1))/((f_(2)-f_(1)))`

`h=(2.480*10^(-19)-1.260*10^(-19))/(738.8*10^(12)-547.5*10^(12))`

`h=6.377*10^(-34)\ J`

Hence, The approximate value of Planck's constant is
`6.377*10^(-34)\ J`