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
External = R
Internal = r
Volume of hemisperical = 2/3 π(R³-r³)
V= 2/3 π(9.1³ - 8.4³)
V= 336.9 cm³
Answer:
The acceleration of the proton is 2.823 x 10¹⁷ m/s²
The acceleration of the electron is 5.175 x 10²⁰ m/s²
Explanation:
Given;
distance between the electron and proton, r = 7 x 10⁻¹⁰ m
mass of proton, = 1.67 x 10⁻²⁷ kg
mass of electron, = 9.11 x 10⁻³¹ kg
The attractive force between the two charges is given by Coulomb's law;
where;
k is Coulomb's constant = 9 x 10⁹ Nm²/c²
Acceleration of proton is given by;
F = ma
Acceleration of the electron is given by;