For an electron to have the same 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?

Wavelength of monochromatic light, λ = 632.8 nm = 632.8 x 10

∴ 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.


For an electron to have the same de broglie wavelength

Hint To answer this question, we should have known about the de broglie's wave equation. Then we can equate the wavelength of the proton and electron and say which of the given statements would satisfy the condition.

Complete step by step answer

Louis de Broglie, a French Physicist, in 1924, proposed the idea that like photons, all material particles such as electron, proton, atom, molecule, a piece of chalk, a piece of stone or iron ball possessed both wave character as well as particle character. He said that the wave associated with a particle is called a matter wave. The wavelength of the wave associated with any material particle was calculated by analogy with photonIf we take a photon, and assume it to have wave character, then the energy of photon is given by $ \Rightarrow E = h\nu {\text{ }} \to {\text{1}}$This is taken from Planck's energy equation.Where,E is the energy of the photonH is the Planck constant $\nu $ is the frequency of the photonAnd$ \Rightarrow E = m{c^2}{\text{ }} \to {\text{2}}$This is taken from the Einstein’s energy equationWhere,m is the mass of photon c is the velocity of light.Equating the equation 1 and 2 we get$ \Rightarrow h\nu = m{c^2}{\text{ }} \to {\text{3}}$We know that frequency is given by$ \Rightarrow \nu = \dfrac{c}{\lambda }$Substituting this in equation 3$ \Rightarrow h\dfrac{c}{\lambda } = m{c^2}$$ \Rightarrow \lambda = \dfrac{h}{{mc}}$This equation gives the wavelength of the photon.We took the particle as a photon for an example. But this equation is applicable for every particle.De Broglie said that the above equation is applicable to any material particle. The mass of the photon is replaced by the mass of the material particle and the velocity “c” of the photon is replaced by the velocity v of the material particle. Thus, for any material particle, the de Broglie’s equation is given by$ \Rightarrow \lambda = \dfrac{h}{{mv}}$ or $ \Rightarrow \lambda = \dfrac{h}{p}{\text{ }} \to {\text{4}}$$\Rightarrow p = mv$$\lambda $ is the wavelength of the particleh is the Planck’s constantm is the massV is the velocityP is the momentumNow in the question it is stated that a proton and electron have the same de broglie wavelength.$ \Rightarrow \lambda = \dfrac{h}{p}{\text{ }}$Rearranging equation, we get$ \Rightarrow p = \dfrac{h}{\lambda }{\text{ }}$So if two particles have same wavelength then their velocity and momentum will be equalIf a proton and electron have the same de broglie wavelength their momentum will be equal.

Hence the correct answer is option (A) Momentum of electron = momentum of proton.


The de broglie's wave equation is only applicable for micro particles. It is not applicable for big particles like cricket balls, because the wavelength of the wave associated with them is too small to be calculated.