Proton and alpha particle have the same de broglie wavelength what is the same for both of them

De-broglie wavelength of the particle is given by, 

`lambda = "h"/"p" = "h"/"mv" = "h"/sqrt(2"mqV")` ;

where, V= Accelerating potential and v is the speed of the particle. 

Given that, the de-broglie wavelength is same for both proton and a-particle.

Charge on α particle = 2c

Mass of α -particle = 4mp

Charge on proton =  qp ;

Mass of α -particle = mP 

`lambda_a = lambda_P`

`=> "h"/sqrt(2"m"_alpha"q"_alpha"V"_alpha) = "h"/sqrt(2 "m"_"P"  "q"_"P" "V"_"P")`

`=> "m"_alpha"q"_alpha"V"_alpha = "m"_"P"  "q"_"P" "V"_"P"`

`=> "V"_"P"/"V"_alpha = ("m"_alpha "q"_alpha)/("m"_"P" "q"_"P") = (4 "m"_"P")/"m"_"P" xx (2"q"_"P")/"q"_"P" = 2/1`

2 : 1 is the required ratio of the accelerating potential.

Also, 

`lambda_"a" = lambda_"p"`

`=> "h"/("m"_alpha "v"_alpha) = "h"/("m"_"P" "V"_"P")`

`=> "V"_"P"/"V"_alpha = "m"_alpha/"m"_"p" = (4"m"_"p")/"m"_"p" = 4/1`

4 : 1 is the required ratio of the speed of proton to speed of alpha-particle.  

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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. 

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  

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

(c) Velocity of hydrogen atom 

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

                    = 0.63 ms-1.