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Question: A particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of t...

A particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is 1.813×10−41.813 \times {10^{ - 4}}. The mass of the particle is (me=9.1×10−31kg)({m_e} = 9.1 \times {10^{ - 31}}kg).
A) 1.67×10−27kg1.67 \times {10^{ - 27}}kg
B) 1.67×10−37kg1.67 \times {10^{ - 37}}kg
C) 1.67×10−19kg1.67 \times {10^{ - 19}}kg
D) 1.67×10−14kg1.67 \times {10^{ - 14}}kg

Explanation

Solution

In order to find the solution of this question, we need to use the de Broglie equation. First of all we need to compare the de Broglie equation for the particle and electron. Then we need to solve the equation, to get the required mass of the particle. We know that the mass .

Complete step by step solution:
As, it is given in the question that the ratio of the wavelength of the particle to the electron, λparticleλelectron=1.813×10−4\dfrac{{{\lambda _{particle}}}}{{{\lambda _{electron}}}} = 1.813 \times {10^{ - 4}}
And the ratios of the velocity of particle to the electron, vparticlevelectron=3\dfrac{{{v_{particle}}}}{{{v_{electron}}}} = 3
Also, mass of the electron, me=9.1×10−31kg{m_e} = 9.1 \times {10^{ - 31}}kg
Now, we know according to the de Broglie equation,
λ=hmv\lambda = \dfrac{h}{{mv}}
At this step, we will equate the wavelength of the particle to the wavelength of the electron according to the de Broglie equation.
So, the ratio becomes,
λparticleλelectron=hmparticle×vparticle×melectron×velectronh\dfrac{{{\lambda _{particle}}}}{{{\lambda _{electron}}}} = \dfrac{h}{{{m_{particle}} \times {v_{particle}}}} \times \dfrac{{{m_{electron}} \times {v_{electron}}}}{h}
⇒1.813×10−4=9.1×10−31kgmparticle×13\Rightarrow 1.813 \times {10^{ - 4}} = \dfrac{{9.1 \times {{10}^{ - 31}}kg}}{{{m_{particle}}}} \times \dfrac{1}{3}
∴mparticle=9.1×10−31kg1.813×10−4×3=1.6852×10−27kg\therefore {m_{particle}} = \dfrac{{9.1 \times {{10}^{ - 31}}kg}}{{1.813 \times {{10}^{ - 4}} \times 3}} = 1.6852 \times {10^{ - 27}}kg
Therefore, the required mass of the particle is 1.6852×10−27kg1.6852 \times {10^{ - 27}}kg.

Hence, option (A), i.e. 1.6852×10−27kg1.6852 \times {10^{ - 27}}kg is the correct answer of the given question.

Note: 1.6852×10−27kg1.6852 \times {10^{ - 27}}kg is the mass of a neutron or proton, so the possibility is that the particle is either a proton or a neutron. The de Broglie wavelength is a wavelength associated with an object and is related to its momentum and mass. The de Broglie equation is given by λ=hmv\lambda = \dfrac{h}{{mv}}. The de Broglie equation is applied to elementary particles, neutral atoms, and molecules. All particles can show wave-like properties. If a particle is significantly larger than its own de Broglie wavelength, or if it is interacting with other objects on a scale significantly larger than its de Broglie wavelength, then its wave-like properties are not noticeable.