When electron and proton are accelerated through the same potential difference which one has greater de Broglie wavelength?

Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.

a.) Find the energy and momentum of each photon in the light beam. 

b.) How many photons per second, on the average arrive at a target irradiated by this beam?

c.) How fast does a hydrogen atom have to travel in order to have the same  momentum as that of the photon?

Given,
Wavelength of monochromatic light, λ = 632.8 nm = 632.8 x 10
–9

 m
∴ Frequency, v = cλ = 3 × 108632.8 × 10-9Hz 

                                 = 4.74 × 1014Hz

(a) Energy of a photon, E = hv

                                       = 6.63 × 10-34× 4.74 × 1014J= 3.14 × 10-19J.

Momentum of each photon, p (momentum) = hλ                            = 6.63 × 10-34632.8 × 10-9                              = 1.05 × 10-27 kg ms-1 

(b) Power emitted, P = 9.42 mW = 9.42 x 10–3 W
Now, P = nE  

This implies, 

n = PE = 9.42 × 10-3W3.14 × 10-19J                = 3 × 1016 photons/sec. 

Thus, these many number of protons arrive at the target.

(c) Velocity of hydrogen atom 


                    = Momentum 'p' of H2 atom (mv)Mass of H2 atom(m) 

                  v = 1.05 × 10-271.673 × 10-27ms-1   

                    = 0.63 ms-1. 


Thus, the hydrogen atom travel at a speed of 0.63 m/s  to have the same  momentum as that of the photon.