A) 1.5 eV, 6 eV, 13.5 eV......, etc.
B) 2 eV, 4 eV, 6 eV ....... etc.
C) 1.5 eV, 3 eV, 6 eV....... etc.
D) 3eV, 5 eV, 7eV......, etc.
Correct Answer: A
Solution :
| de-Broglie waves will resonate with a node at each end of tube. |
| \[\therefore \]\[L=\frac{{{\lambda }_{1}}}{2},\]\[{{\lambda }_{2}},\frac{3}{2}{{\lambda }_{3}}...\frac{n{{\lambda }_{n}}}{2},\]etc |
| So, \[{{\lambda }_{n}},\frac{2L}{n},n=1,2,3...\] As, \[{{\lambda }_{n}}=\frac{h}{{{p}_{n}}}\]and \[{{(KE)}_{n}}=\frac{{{p}^{2}}}{2m}\] |
| We have, \[{{(KE)}_{n}}=\frac{{{n}^{2}}{{h}^{2}}}{8{{L}^{2}}m},n=1,2,3...,\text{etc}\] |
| With \[m=9.1\times {{10}^{-\,31}}\]and \[L=5\times {{10}^{-10}}\,m\] |
| \[{{(KE)}_{n}}=2.4\times {{10}^{-\,19}}{{n}^{2}}J=1.5{{n}^{2}}eV\] |
| \[\therefore \]\[KE=1.5eV,6eV,13.5eV,...,\text{etc}\] |
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