A) We want to find the work function of the potassium. Apply this equation:
E = 1243/λ - Φ
E = energy of photoelectron, λ = incoming light wavelength, Φ = potassium work function
Given values:
E = 2.93eV, λ = 240nm
Plug in and solve for Φ:
2.93 = 1243/240 - Φ
Φ = 2.25eV
B) We want to find the threshold wavelength, i.e. find the wavelength such that the energy E of the photoelectrons is 0eV. Plug in E = 0eV and Φ = 2.25eV and solve for the threshold wavelength λ:
E = 1243/λ - Φ
0 = 1243/λ - Φ
0 = 1243/λ - 2.25
λ = 552nm
C) We want to find the frequency associated with the threshold wavelength. Apply this equation:
c = fλ
c = speed of light in a vacuum, f = frequency, λ = wavelength
Given values:
c = 3×10⁸m/s, λ = 5.52×10⁻⁷m
Plug in and solve for f:
3×10⁸ = f(5.52×10⁻⁷)
f = 5.43×10¹⁴Hz
Answer:
-1
Explanation:
An atom contains a proton, electron and neutron as subatomic particles. The number of protons, which is the positively charged particle (+) is always equal to the number of electrons, which is the negatively charged particle (-) in A NEUTRALLY CHARGED ATOM i.e. no charge.
However, a decrease or increase in charge is dependent on the number of electrons and protons in that atom. Sulfur atom has an electron number of 16 in its neutral atom, however, when it gains one more electron to become 17, it means that the charge has become -1 because the charge = number of protons - number of electrons i.e. 16 - 17 = -1.
I think it #3(not Sure) I think Its because of their long wavelengths that tsunamis behave as shallow-water waves.
The answer is <span>higher than.
</span><span>A sound-producing object is moving toward an observer. The sound the observer hears will have a frequency higher than that actually being produced by the object.
Why?
</span>As the source of the waves is moving toward the observer, each of the successive wave crest<span> is emitted from a position closer to the observer than the previous wave.
Thus each wave takes slightly less time to reach the observer than the previous wave. So, the time between the arrival of successive wave crests at the observer is reduced, increasing the frequency. </span>
Using Kepler's 3rd law which is: T² = 4π²r³ / GM
Solved for r :
r = [GMT² / 4π²]⅓
Where G is the universal gravitational constant,M is the mass of the sun,T is the asteroid's period in seconds, andr is the radius of the orbit.
Change 5.00 years to seconds :
5.00years = 5.00years(365days/year)(24.0hours/day)(6... = 1.58 x 10^8s
The radius of the orbit then is computed:
r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.58 x 10^8s)² / 4π²]⅓ = 4.38 x 10^11m
