When monochromatic light of wavelength 5500 shines on a metal,photo electrons with maximum kinetic energy 1.02 eV are emitted.If the wavelength of the incident light changes to 4800,the maximum kinetic energy of photo electrons becomes 1.35 eV.Calculate the value of the Planck constant based on the information given.
The unit of wavelength is neglected.
The unit of wavelength is neglected.
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The problem represents a system of equations,
((h*ν1 = h*c/λ1 =)) h*c/5500 = E0 + 1.02 eV
h*c/4800 = E0 + 1.35 eV
As c and 1 eV are constants, we have a system of two equations for two unknowns, h and E0.
We can, for example, subtract the two equations to get rid of E0:
h*c/4800 - h*c/5500 = 1.35 eV - 1.02 eV
h*c*(1/4800 - 1/5500) = (1.35 - 1.02) eV
h*c*700/(5500*4800) = 0.33 eV
h = 12445.7 eV/c
More precisely, if the unit of λ is denoted u, we get h in eV c^-1 u.
Using the known values of 1 eV and c, we obtain numerically
h = 6.64 * 10^(-24) J s m^-1
times the unit of wavelength. As a bonus, we can see that both the value and the unit are corrected if u is 10^(-10) m, or one Angstrom.
((h*ν1 = h*c/λ1 =)) h*c/5500 = E0 + 1.02 eV
h*c/4800 = E0 + 1.35 eV
As c and 1 eV are constants, we have a system of two equations for two unknowns, h and E0.
We can, for example, subtract the two equations to get rid of E0:
h*c/4800 - h*c/5500 = 1.35 eV - 1.02 eV
h*c*(1/4800 - 1/5500) = (1.35 - 1.02) eV
h*c*700/(5500*4800) = 0.33 eV
h = 12445.7 eV/c
More precisely, if the unit of λ is denoted u, we get h in eV c^-1 u.
Using the known values of 1 eV and c, we obtain numerically
h = 6.64 * 10^(-24) J s m^-1
times the unit of wavelength. As a bonus, we can see that both the value and the unit are corrected if u is 10^(-10) m, or one Angstrom.