Question

An electron and proton are moving with the same speed. Mass of proton = 1836 times mass
of electron. The ratio of their de-Broglie wavelength will be
(a) 1
(b) 1836
(d) 918
(c) 1836

Answer 1

The answer is attached, hope it helps.

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Answer 2

Answer:

Required ratio of their de-Broglie wavelength is 1836.

Explanation:

Given, An electron and proton are moving with the same speed. Mass of proton = 1836 times mass of electron.

Here we want find ratio of their de-Broglie wavelength.

Now question is what is the concept of de-Broglie wavelength?

De Broglie derived his equation using well established theories through the following series of substitutions:

De Broglie first used Einstein's famous equation relating matter and energy:

$E=m {c}^{2} ......(1)$

with E = energy,

m = mass,

c = speed of light

Using Planck's theory which states every quantum of a wave has a discrete amount of energy given by Planck's equation:

$E = h v .....(2)$

withE = energy,

h = Plank's constant (6.62607 x 10-34 J s),

ν = frequency

Since de Broglie believed particles and wave have the same traits, he hypothesized that the two energies would be equal:

$m {c}^{2} =hν......(3)$

Because real particles do not travel at the speed of light, De Broglie submitted velocity ( v ) for the speed of light ( c ).

$m {v}^{2} =hν......(4)$

Through the equation λ , de Broglie substituted v/λ for ν and arrived at the final expression that relates wavelength and particle with speed.

$m {v}^{2} = \frac{hv}{ λ} .......(5)$

Hence $λ= \frac{hv}{m {v}^{2} } = \frac{h}{mv} ........(6)$

A majority of Wave-Particle Duality problems are simple plug and chug via Equation 6 with some variation of canceling out units.

Given mass of the electron $= m_e$and mass of proton$= m_p$

According to the question,

$m_p = 1836m_e$

From de-Broglie wavelength,

$λ = \frac{h}{p} = \frac{h}{mv}$

$\frac{λ_e}{λ_p} = \frac{m_p}{m_e}$

[v is same]

$= 1836 \frac{m_e}{m_e}$

So,

$\frac{λ_e}{λ_p} = 1836$

Required ratio of their de-Broglie wavelength is 1836.