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Question: The velocity of helium atom is \(1363\ m/s\) ,then the de-Broglie wavelength of helium atom at room ...

The velocity of helium atom is 1363 m/s1363\ m/s ,then the de-Broglie wavelength of helium atom at room temperature is ?
(A) 6.6×10−34 m6.6\times 10^{-34}\ m
(B) 4.39×10−10 m4.39\times 10^{-10}\ m
(C) 7.32×10−14 m7.32\times 10^{-14}\ m
(D) 2.335×10−20 m2.335\times 10^{-20}\ m

Explanation

Solution

The velocity of an atom can be calculated by applying the De Broglie's formula, which relates the wavelength of an atom to its momentum, where momentum is equal to the product of the mass and velocity of the atom.

Complete step by step answer:
The velocity of the helium atom is given as 1363 m/s1363\ m/s. To find the De-Broglie’s wavelength of an atom, let us first recall the De-Broglie’s formula for wavelength, which is given as:
λ=hp\lambda = \dfrac{h}{p} …(i)(i)
Where λ\lambda is the wavelength of an atom;
hh is the Planck's constant;
pp is the momentum of the atom.
Now, p=mass×velocity (v)p = mass\times velocity\ (v)
Therefore, De-Broglie’s formula can also be written as:
λ=hm×v\lambda = \dfrac{h}{m\times v}
⇒λ=hm×v\Rightarrow \lambda = \dfrac{h}{m\times v} …(ii)(ii)
The value of hh is the constant 6.626×10−34 joule.second=6.63×10−34 joule.second6.626\times 10^{-34}\ joule.second = 6.63\times 10^{-34}\ joule.second.
The atomic mass of helium =4 g= 4\ g
Now, the number of atoms present in the atomic mass of an element is equal to Avogadro’s number, which is equal to 6.023×10236.023\times 10^{23}.
⇒\Rightarrow Mass of 6.023×10236.023\times 10^{23} number of helium atoms =4 g=4\ g
⇒\Rightarrow Mass of 11 atom of helium =4 g6.023×1023=\dfrac{4\ g}{6.023\times 10^{23}}
⇒\Rightarrow Mass of 11 atom of helium =0.6641×10−23 g=0.6641\times 10^{-23}\ g
⇒\Rightarrow Mass of 11 atom of helium =0.6641×10−23×1000 kg=0.6641\times 10^{-23}\times 1000\ kg
⇒\Rightarrow Mass of 11 atom of helium =0.6641×10−23 kg=0.6641\times 10^{-23}\ kg
Velocity is given as (v)=1363 m/s(v)=1363\ m/s
Putting the value of hh, mm and vv in equation (ii)(ii), we get;
λ=6.63×10−34 joule.second0.6641×10−23 kg×1363 m/s\lambda = \dfrac{6.63\times 10^{-34}\ joule.second }{0.6641\times 10^{-23}\ kg\times 1363\ m/s}
Now, the value of 1 joule1\ joule in S.I. unit is 1 kg m2 s−21\ kg\ m^2\ s^{-2}.
Therefore, the value of 1 joule.second1\ joule.second in S.I. unit is 1 kg m2 s−2×s1\ kg\ m^2\ s^{-2}\times s.
⇒1 joule.second=1 kg m2 s−1\Rightarrow1\ joule.second = 1\ kg\ m^2\ s^{-1}
⇒λ=6.63×10−34 kg m2 s−10.6641×10−23 kg×1363 ms−1\Rightarrow\lambda = \dfrac{6.63\times 10^{-34}\ kg\ m^2\ s^{-1}}{0.6641\times 10^{-23}\ kg\times 1363\ ms^{-1}}
⇒λ=0.00732×10−11 m\Rightarrow\lambda = 0.00732\times 10^{-11}\ m
⇒λ=7.32×1000×10−11 m\Rightarrow\lambda = 7.32\times 1000\times 10^{-11}\ m
⇒λ=7.32×10−14 m\Rightarrow\lambda = 7.32\times 10^{-14}\ m
Therefore, the velocity of the helium atom is 7.32×10−14 m7.32\times 10^{-14}\ m.

So, the correct answer is Option C.

Note: Avogadro’s Number:
1. Avogadro’s number for an element is the number of atoms present in one mole of the element. Now, one mole of an element is equal to the atomic mass of the element. So, the number of atoms in one mole of an element (for helium, 1 mole=4 g1\ mole = 4\ g) is always equal to the Avogadro’s number.
2. For this question, it must be kept in mind that the velocity of only one helium atom is asked. So, we have to find the mass of one helium atom, and not consider its mass as an entire 4 g4\ g.
Conversion of units:
3. All the values involved in a calculation must be in either S.I. or C.G.S. units. Here, all the options of wavelength are given in m/sm/s, so, we must not forget to convert the mass of a helium atom from gg to kgkg.