Modern Physics 1 Question 33
30. If the mass of the particle is $m=1.0 \times 10^{-30} kg$ and $a=6.6 nm$, the energy of the particle in its ground state is closest to
(2009)
(a) $0.8 meV$
(b) $8 meV$
(c) $80 meV$
(d) $800 meV$
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Solution:
- From Eq. (i) $E=\frac{n^{2} h^{2}}{8 a^{2} m}$
In ground state $n=1$
$$ \therefore \quad E _1=\frac{h^{2}}{8 m a^{2}} $$
Substituting the values, we get
$$ E _1=8 meV $$
