Discussion
A particle of mass m is confined in a three-dimensional cubical box of size L. The number of eigenstates of the Hamiltonian with energy \(\frac{19\hbar^{2}\pi^{2}}{mL^{2}}\) is :
Answer: D
There are 9 eigenstates with this energy.
The levels of a cubical box are
\[ \begin{aligned} E &= \frac{\pi^{2}\hbar^{2}}{2mL^{2}} \\ &\quad\times(n_{x}^{2} + n_{y}^{2} + n_{z}^{2}) \end{aligned} \]
so \(19\hbar^{2}\pi^{2}/mL^{2}\) needs \(n_{x}^{2} + n_{y}^{2} + n_{z}^{2} = 38\) with positive integers. The only ways to write 38 as a sum of three positive squares are
\[ \begin{aligned} 38 &= 36 + 1 + 1 \quad (6, 1, 1) \\ 38 &= 25 + 9 + 4 \quad (5, 3, 2) \end{aligned} \]
(6, 1, 1) can be arranged in 3 ways and (5, 3, 2) in 3! = 6 ways, giving 3 + 6 = 9 states. Stopping at one of the two sets gives 3 or 6, not an offered option; 10 and 4 come from miscounting the arrangements.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development