
A smiley face stress ball having mass 2 g would have highest wavelength of de-Broglie wave associated with it when it travels with velocity
A) 100 m/s
B) 5 m/s
C) 65 m/s
D) 151 m/s
Answer
573.9k+ views
Hint: The de-Broglie equation is, \[\lambda =\dfrac{h}{mv}\]
Where h is Planck's constant, m is the mass in kg and v is velocity in m/s. Plug in the values to this formula to find the desired answer.
Complete Solution :
-First let’s check out the data and requirements of the question. It’s given that we should find out the velocity of the ball that may give the highest de-Broglie wavelength which is associated with the smiley ball.
The de-Broglie equation is given as, \[\lambda =\dfrac{h}{mv}\]
-From this equation if we check how the velocity and wavelength is related to,
\[\lambda \propto \dfrac{1}{v}\]
-Wavelength$\left( \lambda \right)$ is inversely proportional to the velocity and is inversely proportional to mass of the object too.
So, the velocity with lowest value will give the maximum de-Broglie wavelength.
-And by that relation from the above option the velocity of the ball must be 5m/s which is the least value in the given options.
If we have a confusion or doubt in the answer, then we can check the answer of each option by substituting in the above given de-Broglie equation.
Option (A)
\[\lambda =\dfrac{h}{mv}\]=$\dfrac{6.626\times {{10}^{-34}}Js}{2\times {{10}^{-3}}kg\times 100m/s}=3.313\times {{10}^{-33}}m$$\dfrac{6.626\times {{10}^{-34}}Js}{2\times {{10}^{-3}}kg\times 100m/s}=3.313\times {{10}^{-33}}m$
Option (B)
\[\lambda =\dfrac{h}{mv}\]=$\dfrac{6.626\times {{10}^{-34}}Js}{2\times {{10}^{-3}}kg\times 5m/s}=6.626\times {{10}^{-32}}$m
Option (C)
\[\lambda =\dfrac{h}{mv}\] =$\dfrac{6.626\times {{10}^{-34}}Js}{2\times {{10}^{-3}}kg\times 65m/s}=5.0969\times {{10}^{-33}}m$
Option (D)
\[\lambda =\dfrac{h}{mv}\]=$\dfrac{6.626\times {{10}^{-34}}Js}{2\times {{10}^{-3}}kg\times 151m/s}=2.1940\times {{10}^{-33}}m$
Therefore comparing all the answers we can conclude that the least value for velocity will give the maximum value for wavelength.
So, the correct answer is “Option B”.
Note: This question is a tricky question as we may conclude by reading once that we should find the value of velocity from the given data. But the question is pointing to compare the options given and to find the correct value from the options that will give the maximum wavelength.
- By taking just a glimpse of the question, we may get confused that we are not provided with all the valid data and other chances of students solving the equation, is by giving the value of the de-Broglie wavelength of electrons.
- So the question must be read properly in order to decode the real meaning. This question may look like a problem but it is completely a theoretic approach question.
Where h is Planck's constant, m is the mass in kg and v is velocity in m/s. Plug in the values to this formula to find the desired answer.
Complete Solution :
-First let’s check out the data and requirements of the question. It’s given that we should find out the velocity of the ball that may give the highest de-Broglie wavelength which is associated with the smiley ball.
The de-Broglie equation is given as, \[\lambda =\dfrac{h}{mv}\]
-From this equation if we check how the velocity and wavelength is related to,
\[\lambda \propto \dfrac{1}{v}\]
-Wavelength$\left( \lambda \right)$ is inversely proportional to the velocity and is inversely proportional to mass of the object too.
So, the velocity with lowest value will give the maximum de-Broglie wavelength.
-And by that relation from the above option the velocity of the ball must be 5m/s which is the least value in the given options.
If we have a confusion or doubt in the answer, then we can check the answer of each option by substituting in the above given de-Broglie equation.
Option (A)
\[\lambda =\dfrac{h}{mv}\]=$\dfrac{6.626\times {{10}^{-34}}Js}{2\times {{10}^{-3}}kg\times 100m/s}=3.313\times {{10}^{-33}}m$$\dfrac{6.626\times {{10}^{-34}}Js}{2\times {{10}^{-3}}kg\times 100m/s}=3.313\times {{10}^{-33}}m$
Option (B)
\[\lambda =\dfrac{h}{mv}\]=$\dfrac{6.626\times {{10}^{-34}}Js}{2\times {{10}^{-3}}kg\times 5m/s}=6.626\times {{10}^{-32}}$m
Option (C)
\[\lambda =\dfrac{h}{mv}\] =$\dfrac{6.626\times {{10}^{-34}}Js}{2\times {{10}^{-3}}kg\times 65m/s}=5.0969\times {{10}^{-33}}m$
Option (D)
\[\lambda =\dfrac{h}{mv}\]=$\dfrac{6.626\times {{10}^{-34}}Js}{2\times {{10}^{-3}}kg\times 151m/s}=2.1940\times {{10}^{-33}}m$
Therefore comparing all the answers we can conclude that the least value for velocity will give the maximum value for wavelength.
So, the correct answer is “Option B”.
Note: This question is a tricky question as we may conclude by reading once that we should find the value of velocity from the given data. But the question is pointing to compare the options given and to find the correct value from the options that will give the maximum wavelength.
- By taking just a glimpse of the question, we may get confused that we are not provided with all the valid data and other chances of students solving the equation, is by giving the value of the de-Broglie wavelength of electrons.
- So the question must be read properly in order to decode the real meaning. This question may look like a problem but it is completely a theoretic approach question.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

What organs are located on the left side of your body class 11 biology CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

