Find the ratio of the photon wavelength to the de broglie wavelength of the particle. Please help!

The kinetic energy of a particle is equal to the energy of a photon. The particle moves at 4.3% of the speed of light. Find the ratio of the photon wavelength to the de Broglie wavelength of the particle. Take the speed to be non-relativistic.

The energy of a photon is given by:

E = hf = h c / λ, => λ = h c / E = h c / (½ mv²)

The De Broglie wavelength is given by:

λ' = h / p = h / (mv),

The required ratio is:

λ / λ' = [h c / (½ mv²)] / [h / mv] = c / 0.5v

and with v = 0.043 c;

λ / λ' = c / (0.5*0.043c) = 2/0.043 = 46.51

Write down what you know, one statement at a time.

"The kinetic energy of a particle is equal to the energy of a photon."
The energy of a photon is hf = hc/lambda.

Since the particle is non-relativistic, the KE is equal to (1/2)mv^2. So the first piece of information you have is that

(1/2)mv^2 = hc/lambda

"The particle moves at 4.3% of the speed of light."
v = 0.043c

(1/2)m*0.043%2 * c^2 = hc/lambda

This is enough information to solve for lambda in terms of m. Just rearrange the equation.

"Find the ratio of the photon wavelength..."
I just told you how to get an expression for that in terms of m

"...de Broglie wavelength of the particle"
The de Broglie wavelength is h/p = h/mv = h/(m*0.043c)

Divide the first wavelength by the second wavelength as requested. Simplify. I believe you'll find the m's cancel out.