The threshold wavelength for photoelectric emission in metal is 600 nm.Find the maximum kinetic energy of the electrons emitted when it is exposed to the wavelength 200 nm
KE of emission=energyIncident-workfunction
workfuncion=hc/600nm solve that in joules
then energy incident=hc/200nm
KE=hc(1/200-1/600)1/nm=hc/1E-9 (4 /600)=4*E-9*hc/600
To find the maximum kinetic energy of the electrons emitted, we will use the equation:
E = hc / λ
where:
- E is the energy of a photon,
- h is the Planck's constant (6.626 x 10^-34 J.s),
- c is the speed of light (3.00 x 10^8 m/s),
- λ is the wavelength of the light.
First, let's calculate the energy of a photon with a wavelength of 200 nm:
λ = 200 nm = 200 x 10^-9 m
E = (6.626 x 10^-34 J.s)(3.00 x 10^8 m/s) / (200 x 10^-9 m)
E = 9.939 x 10^-19 J
Now, we will calculate the energy of a photon with the threshold wavelength of 600 nm:
λ = 600 nm = 600 x 10^-9 m
E_threshold = (6.626 x 10^-34 J.s)(3.00 x 10^8 m/s) / (600 x 10^-9 m)
E_threshold = 3.313 x 10^-19 J
The difference between these two energies will give us the maximum kinetic energy of the emitted electron:
K.E. = E - E_threshold
K.E. = (9.939 x 10^-19 J) - (3.313 x 10^-19 J)
K.E. = 6.626 x 10^-19 J
Therefore, the maximum kinetic energy of the electrons emitted is 6.626 x 10^-19 J.
To find the maximum kinetic energy of the electrons emitted when exposed to a wavelength of 200 nm, we need to use the equation for the energy of a photon:
E = hc/λ,
where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J*s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of light.
First, we need to find the energy of a photon with a wavelength of 200 nm. Plugging the values into the equation:
E = (6.626 x 10^-34 J*s * 3.00 x 10^8 m/s) / (200 x 10^-9 m)
= 9.939 x 10^-19 J
Next, we need to find the threshold energy for photoelectric emission, which is given by multiplying the threshold wavelength by the energy of a photon at that wavelength:
Eth = hc/λth,
where λth is the threshold wavelength of 600 nm. Plugging the values into the equation:
Eth = (6.626 x 10^-34 J*s * 3.00 x 10^8 m/s) / (600 x 10^-9 m)
= 3.313 x 10^-19 J
Finally, we can calculate the maximum kinetic energy of the electrons emitted by subtracting the threshold energy from the energy of the photon:
KEmax = E - Eth
= (9.939 x 10^-19 J) - (3.313 x 10^-19 J)
= 6.626 x 10^-19 J
Therefore, the maximum kinetic energy of the electrons emitted when exposed to a wavelength of 200 nm is 6.626 x 10^-19 J.