Question

(a) A light beam of wavelength 7.0 x 10^-7 m and intensity 435 W/m2 shines on a target of area 3.8 m2. How many photons per second are striking the target? Answer: ______________.___x 10 photons/sec

(b) A particle of matter is moving with a kinetic energy of 10 eV. Its de Broglie wavelength is 7.3 x 10^-10 m. What is the mass of the particle? Answer: _________________.___x 10 kg

(c) Two spaceships are traveling with a relative velocity of 2.9 x 10⁸ m/s. Both carry identical clocks. According to the captain of each ship, the other captain’s clock ticks more slowly than his own. By what factor do the two clocks disagree? Answer: ______________.___

Answer 1

a) The power of the light is
P = I A
P = 435 W/m² x 3.8 m²
P = 1653 W
The energy transferred in one second is
E_t = 1653 J
Energy of one photon
E = h c /
Where is the wavelength of the light used
E = 6.63 x 10^-34 x 3 x10⁸ m/s / 7 x 10^-7 m
E = 2.84 x 10^-19 J
Thus the number of photons in the light beam hitting the target in one second is
n = E_t / E = 1653 J / 2.84 x 10^-19 J
n = 5.820 x 10²¹
b) The kinetic energy of the particle is 10 eV
The momentum of the particle is
p = h /
Where is the de Broglie wavelength
p = 6.63 x 10^-34 J s / 7.3 x 10^-10 m
p = 9.08 x 10^-25 kgm/s
The kinetic energy K = p² /2 m
m = p²/ 2 K
m = (9.08 x 10^-25)² / (2 x 10 x 1.6 x 10^-19)
m = 2.58 x 10^-31 kg
c) The dilated time will be
t = t₀ / sqrt (1- v² / c²)
The factor is
  = 1 / sqrt(1 - v² /c²)
  = 1 /  sqrt(1 - (2.9 x 10⁸)² /(3 x 10⁸)²)
  = 1.482