Modern Physics Ques 41
- 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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Answer:
Correct Answer: 41.(b)
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 $
BETA
