Question

Consider an isolated hydrogen atom of mass 1.66 x 10-27 kg. (a) Find the gravitational force on this hydrogen atom near the surface of the earth (assume that at sea level the gravitational acceleration constant g= 9.8 m/s2 ). (b) Let an upwardly directed laser beam emitting 1-eV photons be forced in such a way that the full momentum of each of its photons is transferred to the atom. Find the average upward force on the atom provided by one photon striking each second. (c) Find the number of photons that must strike the atom per second, and the corresponding optical power, for it not to fall under the effect of gravity, given idealized conditions in vacuum. (d) How many photons per second would be required to keep the atom from falling if it were perfectly reflecting?

Answer 1

a)

m = mass of atom = 1.66 x 10^-27 kg

g = acceleration due to gravity = 9.8 m/s²

gravitational force on hydrogen atom is given as

F_g = mg

F_g = (1.66 x 10^-27) (9.8)

F_g = 1.63 x 10^-26 N

b)

E = energy of each photon = 1 eV = 1.6 x 10^-19 J

c = speed of photon = 3 x 10⁸ m/s

P = momentum of photon

Momentum of each photon is given as

P = E/c

P = (1.6 x 10^-19 ) /(3 x 10⁸)

P = 5.3 x 10^-28

F = average upward force = ?

t = time = 1 sec

Using the equation for impulse

F t = P

F (1) = 5.3 x 10^-28

F = 5.3 x 10^-28 N

c)

n = number of photons per sec

n = F_g /F

n = (1.63 x 10^-26)/( 5.3 x 10^-28)

n = 31

d)

when perfectly reflecting

P' = 2E/c = 2 x 5.3 x 10^-28 = 10.6 x 10^-28

Using the equation for impulse

F' t = P'

F' (1) = 10.6 x 10^-28

F' = 10.6 x 10^-28 N

n' = number of photons per sec

n' = F_g /F'

n' = (1.63 x 10^-26)/( 10.6 x 10^-28)

n' = 15