When photons with a wavelength of 232 nm are incident on a metal with a work function =1.78 eV., the maximum kinetic energy (in eV to two decimal places) of photoelectrons emitted are
Question
When photons with a wavelength of 232 nm are incident on a metal with a work function =1.78 eV., the maximum kinetic energy (in eV to two decimal places) of photoelectrons emitted are
Solution
To solve this problem, we need to use the photoelectric effect equation:
KE_max = hf - Φ
where KE_max is the maximum kinetic energy of the photoelectrons, h is Planck's constant, f is the frequency of the incident light, and Φ is the work function of the metal.
First, we need to convert the wavelength of the incident light to frequency using the equation:
f = c/λ
where c is the speed of light (3.00 x 10^8 m/s) and λ is the wavelength. However, the wavelength is given in nanometers, so we need to convert it to meters:
232 nm = 232 x 10^-9 m
Now we can calculate the frequency:
f = (3.00 x 10^8 m/s) / (232 x 10^-9 m) = 1.29 x 10^15 Hz
Next, we need to convert the work function from eV to Joules, since the other quantities in the equation are in SI units. 1 eV = 1.60 x 10^-19 J, so:
Φ = 1.78 eV = 1.78 x 1.60 x 10^-19 J = 2.85 x 10^-19 J
Now we can calculate the maximum kinetic energy. Planck's constant h = 6.63 x 10^-34 J.s, so:
KE_max = (6.63 x 10^-34 J.s)(1.29 x 10^15 Hz) - 2.85 x 10^-19 J = 1.70 x 10^-19 J
Finally, we need to convert this energy back to eV:
KE_max = 1.70 x 10^-19 J / (1.60 x 10^-19 J/eV) = 1.06 eV
So, the maximum kinetic energy of the photoelectrons is approximately 1.06 eV.
Similar Questions
The maximum kinetic energy of emitted photoelectrons, from different metalsurfaces, depends upon:A. Intensity of incident light only ⃝B. Frequency of the incident light only ⃝C. Nature of metal surface ⃝D. Both frequency of incident light and nature of metal surface
Radiations of frequency, (v) are incident on a photosensitive metal. The maximum kineticenergy of photoelectrons is E. When the frequency of the incident radiations is doubled, then whatis the maximum kinetic energy of the photoelectrons?
The work function for a metal is 4.31 eV. Ultraviolet light of wavelength 125 nm is incident on the clean surface of this metal. What is the photon energy of the ultraviolet light in units of eV?Answer for part 1What is the maximum kinetic energy (in eV) of the ejected photoelectrons?Answer for part 2What stopping voltage would be required to arrest the current of photoelectrons?
For the photoelectric effect, the maximum kinetic energy (Ek) of the photoelectrons is plotted against the frequency (v) of the incident photons as shown in figure. The slope of the graph gives
The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential in volts is :
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.