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Question: If the mass of the particle is \[m=1.0\times {{10}^{-30}}kg\] and \(a=6.6nm\), the energy of the par...

If the mass of the particle is m=1.0×10−30kgm=1.0\times {{10}^{-30}}kg and a=6.6nma=6.6nm, the energy of the particle in its ground state is closest to
A.0.8meVA.0.8meV
B.8meVB.8meV
C.80meVC.80meV
D.800meVD.800meV

Explanation

Solution

We are going to use the Bohr’s model of Atom to solve this problem. We know that the energy of the particle is related to its quantum number. The particle of mass mm is related to its linear momentum with the relationE=p22mE=\dfrac{{{p}^{2}}}{2m}.

Formula used:
We are going to use the following relation to solve the above given problem:-
E=n2h28a2mE=\dfrac{{{n}^{2}}{{h}^{2}}}{8{{a}^{2}}m}.

Complete answer:
We are using the following given formula to solve the above given problem:-
E=n2h28a2mE=\dfrac{{{n}^{2}}{{h}^{2}}}{8{{a}^{2}}m}…………………. (i)(i)
Where EE energy of the particle, nn represents principle quantum number, hh is Planck’s constant, mm is the mass of the particle and aa is the acceleration of the particle.
From the problem we have,
n=1n=1(Particle is in ground state)
h=6.6×10−34h=6.6\times {{10}^{-34}}
a=6.6nm=6.6×10−9ma=6.6nm=6.6\times {{10}^{-9}}m
m=m=mass of the particle.
Putting values in (i)(i)we get
E=(1)2×(6.6×10−34)28×(6.6×10−9)2×1×10−30×1.6×10−19E=\dfrac{{{(1)}^{2}}\times {{(6.6\times {{10}^{-34}})}^{2}}}{8\times {{(6.6\times {{10}^{-9}})}^{2}}\times 1\times {{10}^{-30}}\times 1.6\times {{10}^{-19}}}
We have multiplied the denominator by 1.6×10−19C1.6\times {{10}^{-19}}C because we need the answer in electron volts.
Solving further we get,
E=8×10−3eVE=8\times {{10}^{-3}}eV
E=8meVE=8meV

So, the correct answer is “Option B”.

Additional Information:
The electron rotates about the nucleus in certain stationary circular orbits, for which the angular momentum of electron about the nucleus is an integral multiple of h2π=η\dfrac{h}{2\pi }=\eta , where h is planck's constant. The atom radiates energy when the electron jumps from one allowed stationery state to another. It could not explain the spectra of atoms containing more than one electron.

Note:
It should be noted that these relations are applicable to hydrogen like species. Energy of electrons increases (decreases in negative) as the orbit becomes higher. Value of nn should be used correctly. We need to find the answer in electron volts, therefore we multiply the denominator with 1.6×1019C1.6\times {{10}^{19}}C otherwise we don’t need to do that.