Answer:a) 492 nm; b) 1.78 eV
Explanation: In order to solve this problem we have to use the photoelectric energy balance givenb by:
h*ν=Ek+W where h is the Placnk constant, ν is the frequency of the radiation, Ek is the kinetic energy of the realised electrons and W is the work funcion of the incident metal.
Then we have:
W=h*ν-Ek= h*c/λ-Ek=1240 eV.nm/176 nm=2.52 eV
so the maximun wavelength to realese the electrons is when Ek=0 then
W=h*c/λmax so λmax= h*c/W= 1240/2.52=492.06 nm
Finally if we use 288 nm to realease the electrons, then Ek of the emitted electron from the metal is:
from h*ν=Ek+W we have:
Ek=h*c/λ-W= (1240/288) eV*nm-2.52 eV=1.78 eV
Answer
30 mi/gal * (1.61 km / mi ) / (3.78 L/gal) = 30 * .426 km/L = 12.8 km/L
The energy carried by the incident light is
where h is the Planck constant and f is the frequency of the light. The threshold frequency is the frequency that corresponds to the minimum energy needed to eject the electrons from the metal, so if we substitute the threshold frequency in the formula, we get the minimum energy the light must have to eject the electrons:
Different:
1. number of moons
2. size of rings
3. mass
4. temperature
5. type of rings
i might be wrong though.