Cobalt-60 is an artificial radioisotope that is produced in a nuclear reactor for use as a gamma-ray source in the treatment of certain types of cancer. If the wavelength of the gamma radiation from Co-60 is 2.00 x 10-12 m, calculate the energy of a photon of this radiation.

E = 9.93 x 10^-14 J

Explanation:

Wavelength, λ = 2.00 x 10-12 m

Energy, E = ?

Relationship between both quantities is given as;

E = hc/λ

where;

c = speed of light = 2.998  ×  10 8  m⋅s −1

h = Plank's constant = 6.626*10-34 Js

Solving for E;

E = 6.626*10^-34 * 2.998  ×  10^ 8 / 2.00 x 10^-12

E = 9.93 x 10^-14 J

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j