Answer
Hint: The de Broglie wavelength is basically the wavelength associated with a moving particle/matter. An alpha particle is basically He2+ ion, i.e consisting of two neutrons and two protons.
Formula Used: $\lambda=\dfrac{h}{mv}$
Complete step by step answer: As we all know that the de-broglie equation states that matter can act as waves not much like light and some radiation will behave like both particles and waves. De-broglie equation helps us to imagine that matter is also having wavelength.
We have studied that the de-broglie equation is given by:$\lambda = \dfrac{h}{{mv}}$ …… (I)Here $\lambda $ is the De-broglie wavelength, $h$ is the Planck's constant, $m$ is the mass of the particle and $v$ is the velocity of the particle.So, If the de-broglie wavelength for both the proton and alpha particle is same then the equation (I) can be written as,$ \Rightarrow {\lambda _1} = {\lambda _2}$$ \Rightarrow {\left( {\dfrac{h}{{mv}}} \right)_1} = {\left( {\dfrac{h}{{mv}}} \right)_2}$ …… (II)Here the subscripts 1 and 2 represent the proton and the alpha particle. $m$ represents the mass and $v$ represents the velocity. Also it is given to us in the question that,${m_2} = 4{m_1}$Here ${m_1}$ is the mass of the proton and ${m_2}$ is the mass of the alpha particle. Now the equation (II) can be written by substituting ${m_2} = 4{m_1}$ as,\[ {m_1}{v_1} = {m_2}{v_2} \\ \Rightarrow {m_1}{v_1} = 4{m_1}{v_2} \\ \Rightarrow {v_1} = 4{v_2} \\ \therefore \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{4}{1} \\ \] Therefore, the velocity of the proton is four times the mass of the alpha particle. Therefore, the correct option is (D).Note: As we all know that the concept regarding wave particle duality is well explained by matter waves. Matter of all kinds can exhibit a wave like behaviour. A best example for it is that a beam of electrons is just diffracted like the beam of light or water wave. But in general the wavelength is very small to impact a great extent on us for our day to day life. We mean that the matter waves related to tennis ball, other objects are not relevant.
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
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.