# Consider a particle in an infinite deep potential well. choose the correct pairs of energy in 1st and 2nd excited state

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Consider a particle in an infinite deep potential well. choose the correct pairs of energy in 1st and 2nd excited state
1. $\frac{h^2}{8mL^2} , \frac{4h^2}{8mL^2}$
2. $\frac{4h^2}{8mL^2} ,\frac{9h^2}{8mL^2}$
3. $\frac{h^2}{8mL^2},\frac{2h^2}{8mL^2}$
4. $\frac{2h^2}{8mL^2}, \frac{3h^2}{8mL^2}$

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Correct Answer - Option 2 : $\frac{4h^2}{8mL^2} ,\frac{9h^2}{8mL^2}$

Concept:

• Thus the normalized wave function of a particle in an infinite deep potential well is$\psi = \sqrt{\frac{2}{L}} sin(\frac{n\pi x}{L})$
• The energy of the particle in an infinitely deep potential well can be mathematically expressed as, $E_n = \frac{n^2 \pi^2\hbar^2}{2 m L^2}$  where n is a principal quantum number (n=1,2,3,4,......), L is the length of the box.
• $\because \hbar = h/2\pi \\ \;energy\; can \; be \; wrtiiten\; as \;\\ E = \frac{n^2 h^2}{8mL^2}$

Explanation:

for the 1st excited state value of n = 2 thus energy is

$E = \frac{4 h^2}{8mL^2}$

for 2nd excited state value of n=3 thus energy is

$E = \frac{9 h^2}{8mL^2}$

hence option 2 is correct