Answer:
(a) The energy of the photon is 1.632 x 
J.
(b) The wavelength of the photon is 1.2 x 
m.
(c) The frequency of the photon is 2.47 x 
Hz.
Explanation:
Let;
= -13.60 ev
= -3.40 ev
(a) Energy of the emitted photon can be determined as;
- = -3.40 - (-13.60)
= -3.40 + 13.60
= 10.20 eV
= 10.20(1.6 x 
)
- = 1.632 x Joules
The energy of the emitted photon is 10.20 eV (or 1.632 x Joules).
(b) The wavelength, λ, can be determined as;
E = (hc)/ λ
where: E is the energy of the photon, h is the Planck's constant (6.6 x 
Js), c is the speed of light (3 x 
m/s) and λ is the wavelength.
10.20(1.6 x ) = (6.6 x * 3 x )/ λ
λ = 
= 1.213 x
Wavelength of the photon is 1.2 x m.
(c) The frequency can be determined by;
E = hf
where f is the frequency of the photon.
1.632 x = 6.6 x x f
f = 
= 2.47 x Hz
Frequency of the emitted photon is 2.47 x Hz.
The vector sum of forces acting on a non-accelerating object equals zero.
equation form: ΣF = 0
<span>The electric force is given by:
F = [ k*(q1)*(q2) ] / d^2
F = Electric force
k = Coulomb's constant
q1 = Charge of one proton
q2 = Charge of second proton
d = Distance between centers of mass
Values:
F = unknown
k = 8.98E 9 N-m^2/C^2
q1 = 1.6E-19
q2 = 1.6E-19
d = 1.0E-15 m
Insert values into F = [ k*(q1)*(q2) ] / d^2
F = [ (8.98E 9 N-m^2/C^2) * (1.6E-19) * (1.6E-19) ] / (1.0E-15 m)^2
F = </span>229.888 N
answer
the electric force of repulsion between nuclear protons is 229.888 N
The power dissipated is simply V^2/R
where V = 120 volts RMS
and R = 60 Ω
