
If He atom is moving with $2 \times {10^2}m/s$ then calculate the associated wavelength with this He atom?
A) $4.79\mathop A\limits^ \circ $
B) $5.21\mathop A\limits^ \circ $
C) $4.2\mathop A\limits^ \circ $
D) $3.9\mathop A\limits^ \circ $
Answer
581.1k+ views
Hint: The wavelength associated with the particles is called de Broglie wavelength, so we can use de Broglie wavelength equation to calculate the wavelength associated with He atom i.e. $\lambda = \dfrac{h}{{mv}}$
Where $h = $ planck’s constant, m= mass of particle, and v=velocity of particle
Complete step by step answer:
Step1: Writing down what is given in the question and what is need to calculate-
Velocity of the He atom, $v = 2 \times {10^2}m/s$,
Mass of He atom =4 units=$4 \times 1.67 \times {10^{ - 27}}kg$
Planck’s constant $h = 6.626 \times {10^{ - 34}}Js$
Wavelength $\lambda = ?$(need to calculate)
Step2: Now apply the de Broglie equation which is given by -
$\lambda = \dfrac{h}{{mv}}$
Where, $h = $ planck’s constant, m= mass of He atom, and v=velocity of He atom
Substitute all the data in above equation,
$\lambda = \dfrac{{6.626 \times {{10}^{ - 34}}}}{{4 \times 4 \times 1.67 \times {{10}^{ - 27}} \times 2 \times {{10}^2}}}$
Step3: Now simplify the expression to calculate the value of wavelength.
On simplifying we get,
$\lambda = 0.4759 \times {10^{ - 9}}m$
$\left( {\because 1\mathop A\limits^ \circ = {{10}^{ - 10}}m} \right)$
$ \Rightarrow \lambda = 4.759\mathop A\limits^ \circ $
Therefore the required wavelength of the He atom is $ 4.759\mathop A\limits^ \circ $ . So option (A) is correct.
Additional Information:
The physical significance of de Broglie wavelength according to wave-particle duality is a wavelength manifested in all the objects in quantum mechanics which determines the probability density of finding the object at a given point of the configuration space. De Broglie waves are also called as the matter waves because they are associated with the matter particles.
Also, De-Broglie was awarded a Nobel prize for this discovery of matter waves.
Note:
Always keep in mind the unit conversions of the dimensions and physical quantity. Moreover many times Planck’s constant is not given so try to learn the values of such constants.
Commit to your memory: The value of planck’s constant (h): $h = 6.626 \times {10^{ - 34}}Js$.
Where $h = $ planck’s constant, m= mass of particle, and v=velocity of particle
Complete step by step answer:
Step1: Writing down what is given in the question and what is need to calculate-
Velocity of the He atom, $v = 2 \times {10^2}m/s$,
Mass of He atom =4 units=$4 \times 1.67 \times {10^{ - 27}}kg$
Planck’s constant $h = 6.626 \times {10^{ - 34}}Js$
Wavelength $\lambda = ?$(need to calculate)
Step2: Now apply the de Broglie equation which is given by -
$\lambda = \dfrac{h}{{mv}}$
Where, $h = $ planck’s constant, m= mass of He atom, and v=velocity of He atom
Substitute all the data in above equation,
$\lambda = \dfrac{{6.626 \times {{10}^{ - 34}}}}{{4 \times 4 \times 1.67 \times {{10}^{ - 27}} \times 2 \times {{10}^2}}}$
Step3: Now simplify the expression to calculate the value of wavelength.
On simplifying we get,
$\lambda = 0.4759 \times {10^{ - 9}}m$
$\left( {\because 1\mathop A\limits^ \circ = {{10}^{ - 10}}m} \right)$
$ \Rightarrow \lambda = 4.759\mathop A\limits^ \circ $
Therefore the required wavelength of the He atom is $ 4.759\mathop A\limits^ \circ $ . So option (A) is correct.
Additional Information:
The physical significance of de Broglie wavelength according to wave-particle duality is a wavelength manifested in all the objects in quantum mechanics which determines the probability density of finding the object at a given point of the configuration space. De Broglie waves are also called as the matter waves because they are associated with the matter particles.
Also, De-Broglie was awarded a Nobel prize for this discovery of matter waves.
Note:
Always keep in mind the unit conversions of the dimensions and physical quantity. Moreover many times Planck’s constant is not given so try to learn the values of such constants.
Commit to your memory: The value of planck’s constant (h): $h = 6.626 \times {10^{ - 34}}Js$.
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