
The de Broglie wavelength associated with a ball of mass \[200\]g and moving at a speed of \[5\]meters / hour , is of the order of (h\[ = 6.625 \times {10^{-34}}\]):
A. $10^{-15}$
B. $10^{-20}$
C. $10^{-25}$
D. $10^{-30}$
Answer
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Hint: As we know that the de Broglie equation is developed by Louis de Broglie . The de Broglie derived a connection between the wavelength and momentum of matter. The de Broglie wavelength is inversely proportional to its momentum.
Complete answer: or Complete step by step answer:
De Broglie wavelength: The de Broglie wavelength is an important concept when studying quantum mechanics. The wavelength \[(\lambda )\] that is associated with an object with respect to its speed and mass is known as the de Broglie wavelength. A particle de Broglie wavelength is usually inversely proportional to its force.
De Broglie Equation: the wave mechanic postulates that a particle of mass ‘m’ rotating at a velocity ‘v’ will have the properties of the wavelength \[\dfrac{h}{{mv}}\] (de Broglie wavelength), where ‘h’ is the plank constant. We use de Broglie equation to measure the wavelength, momentum, frequencies or kinetic energy of particles.
Mass: Mass is the amount of matter in an object.
Speed: The speed of an object is the result of a change in its position.
Velocity: Velocity is a displacement of an object in unit time.
According to the question, the de Broglie wavelength is connected with a ball of mass \[200g\] and moving at a speed of \[5meters/hour\].
As we known that the De Broglie equation is;
\[\lambda = \dfrac{h}{{mv}}\]
Where, \[\lambda \]is the de Broglie wavelength
\[h\]is the Planck’s constant, ‘\[h\]’= \[6.626 \times {10^{ - 34}}Js\]
m is the mass of object, ‘m’ = \[200g = 0.2kg\]
v is velocity ‘v’ = \[5m/h = 5/3600m/s\]
According to the formula of de Broglie equation;
\[\lambda = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3600}}{{0.2 \times 5}}\]
Where the value of mass of object and velocity is given in question.
Thus, \[\lambda = 2.385 \times {10^{ - 30}}m\]
Hence, option (D) is a correct answer.
Note:
The de Broglie equation states that a substance can act as waves, like light and radiation, which also behave as waves and particles.
The de Broglie equations concern the wavelength to the momentum ’p’ and frequency ‘f’ to the total energy ‘E’ of a free particle.
Complete answer: or Complete step by step answer:
De Broglie wavelength: The de Broglie wavelength is an important concept when studying quantum mechanics. The wavelength \[(\lambda )\] that is associated with an object with respect to its speed and mass is known as the de Broglie wavelength. A particle de Broglie wavelength is usually inversely proportional to its force.
De Broglie Equation: the wave mechanic postulates that a particle of mass ‘m’ rotating at a velocity ‘v’ will have the properties of the wavelength \[\dfrac{h}{{mv}}\] (de Broglie wavelength), where ‘h’ is the plank constant. We use de Broglie equation to measure the wavelength, momentum, frequencies or kinetic energy of particles.
Mass: Mass is the amount of matter in an object.
Speed: The speed of an object is the result of a change in its position.
Velocity: Velocity is a displacement of an object in unit time.
According to the question, the de Broglie wavelength is connected with a ball of mass \[200g\] and moving at a speed of \[5meters/hour\].
As we known that the De Broglie equation is;
\[\lambda = \dfrac{h}{{mv}}\]
Where, \[\lambda \]is the de Broglie wavelength
\[h\]is the Planck’s constant, ‘\[h\]’= \[6.626 \times {10^{ - 34}}Js\]
m is the mass of object, ‘m’ = \[200g = 0.2kg\]
v is velocity ‘v’ = \[5m/h = 5/3600m/s\]
According to the formula of de Broglie equation;
\[\lambda = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3600}}{{0.2 \times 5}}\]
Where the value of mass of object and velocity is given in question.
Thus, \[\lambda = 2.385 \times {10^{ - 30}}m\]
Hence, option (D) is a correct answer.
Note:
The de Broglie equation states that a substance can act as waves, like light and radiation, which also behave as waves and particles.
The de Broglie equations concern the wavelength to the momentum ’p’ and frequency ‘f’ to the total energy ‘E’ of a free particle.
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