JEE Advance - Physics (2009 - Paper 1 Offline - No. 14)
If the mass of the particle is $$m=1.0\times10^{-30}$$ kg and $$a=6.6$$ nm, the energy of the particle in its ground state is closest to
0.8 meV
8 meV
80 meV
800 meV
Explanation
For ground state, $$n = 1$$. Therefore,
$${E_1} = {{{h^2}} \over {8m{a^2}}} = \left[ {{{{{(6.6 \times {{10}^{ - 34}})}^2}} \over {8 \times {{10}^{ - 30}} \times {{(6.6 \times {{10}^{ - 9}})}^2}}}} \right]$$ J
Therefore, $${{{E_1}} \over e} = \left( {{{{E_1}} \over {1.6 \times {{10}^{ - 19}}}}} \right)$$ meV = 8 meV
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