Question

A hydrogen atom, initially at rest, absorbs an ultraviolet photon with a wavelength of λ=177.9nm.

Part A:

What is the atom's final speed if it now emits an identical photon in a direction that is perpendicular to the direction of motion of the original photon?

Express your answer to three significant figures and include appropriate units.

Part B:

What is the atom's final speed if it now emits an identical photon in a direction that is opposite to the direction of motion of the original photon?

Express your answer to three significant figures and include appropriate units.

Answer 1

given

= 177.9 nm

= 177.9 x 10^-9 m

a )

using equation P = h /

= 6.627 x 10^-34 / 177.9 x 10^-9

P = 3.7251 x 10^-27 kg-m/sec

initial momentum P_x = 3.7251 x 10^-27 i

final momentum P_y = 3.7251 x 10^-27 j

change in momentum = 3.7251 x 10^-27 i - 3.7251 x 10^-27 j

magnitude = ( (3.7251 x 10^-27 )² + ( 3.7251 x 10^-27 )² )^1/2

= 5.268 x 10^-27 N-sec

here we have change in momentum is = m v

m v = 5.268 x 10^-27

v = 5.268 x 10^-27 / 1.67 x 10^-27

so the atom's final speed if it now emits an identical photon in a direction

that is perpendicular to the direction of motion of the original photon is v = 3.1544 m/sec

b )

in opposite direction = ( m v- ( - mv ) )

= 2 m v

= 2 x 3.7251 x 10^-27

then now

2 x 3.7251 x 10^-27 = 1.67 x 10^-27 x v

2 x 3.7251 = 1.67 x v

v = 2 x 3.7251 / 1.67

so the atom's final speed if it now emits an identical photon in a direction

that is opposite to the direction of motion of the original photon v = 4.4933 m/sec