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← IIT-JEE 2009 · Paper 1
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IIT-JEE 2009 · Paper 1, Physics question 57

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Paragraph for Question Nos. 56 to 58

When a particle is restricted to move along xx-axis between x=0x=0 and x=ax=a, where aa is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x=0x=0 and x=ax=a. The wavelength of this standing wave is related to the linear momentum pp of the particle according to the de Broglie relation. The energy of the particle of mass mm is related to its linear momentum as E=p22mE=\dfrac{p^2}{2m}. Thus, the energy of the particle can be denoted by a quantum number 'nn' taking values 1, 2, 3, … (n=1n=1, called the ground state) corresponding to the number of loops in the standing wave.

Use the model described above to answer the following three questions for a particle moving in the line x=0x=0 to x=ax=a. Take h=6.6×10−34h=6.6\times10^{-34} J s and e=1.6×10−19e=1.6\times10^{-19} C.

If the mass of the particle is m=1.0×10−30m=1.0\times10^{-30} kg and a=6.6a=6.6 nm, the energy of the particle in its ground state is closest to

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Answer and worked solution

Official answer: B

Ground state.
E1=h28ma2=(6.6×10−34)28(10−30)(6.6×10−9)2=10−508×10−30=1.25×10−21 J.E_1=\frac{h^2}{8ma^2}=\frac{(6.6\times10^{-34})^2}{8(10^{-30})(6.6\times10^{-9})^2}=\frac{10^{-50}}{8\times10^{-30}}=1.25\times10^{-21}\ \text{J}.

In eV. 1.25×10−211.6×10−19≈7.8×10−3\dfrac{1.25\times10^{-21}}{1.6\times10^{-19}}\approx7.8\times10^{-3} eV ≈ 8 meV.

(B)