What's the quantum number for a particle in an infinite square well if the particle's energy is 64 times the ground-state energy?

8

Explanation:

The energy levels is given as

E(n) = n² * h² / ( 8 * m * L²), where

n = 1, 2, 3, 4,... etc

At ground state energy (n = 1), therefore is E(g) = h² / (8 * m * L²).

We can then say that

E(n) = n² * Eg

Therefore, to have E = 64 Eg, we must have n² = 64. And for n² to be equal to 64, we find the square root of n

n = √64

n = 8

Essentially, the needed quantum number is 8

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2