the work function of zinc is 3.6ev what is maximum enegy photoelectron by ultraviolet wave of light 3000angstrome
To find the maximum energy of a photoelectron emitted by ultraviolet light with a wavelength of 3000 Å (angstroms) on zinc, we can make use of the relationship between energy (E) and wavelength (λ) in terms of the photon energy equation.
The photon energy equation is given by:
E = hc / λ
Where:
E is the energy of one photon (in joules)
h is Planck's constant (6.626 x 10^-34 J·s)
c is the speed of light (2.998 x 10^8 m/s)
λ is the wavelength of the light (in meters)
First, we need to convert the given wavelength from angstroms (Å) to meters (m). Since 1 Å = 1 x 10^-10 m, the conversion is:
λ = 3000 Å x (1 x 10^-10 m / 1 Å) = 3 x 10^-7 m
Next, we can substitute the values into the photon energy equation:
E = (6.626 x 10^-34 J·s) * (2.998 x 10^8 m/s) / (3 x 10^-7 m)
E ≈ 6.976 x 10^-19 J
To convert the energy from joules (J) to electron volts (eV), we can use the conversion factor:
1 eV = 1.602 x 10^-19 J
Maximum energy of the photoelectron = (6.976 x 10^-19 J) / (1.602 x 10^-19 J/eV)
Maximum energy ≈ 4.35 eV
Therefore, the maximum energy of the photoelectron emitted by ultraviolet light with a wavelength of 3000 Å (angstroms) on zinc is approximately 4.35 eV.