DeBroglie wavelength in infinite potential well

[lowcard2]

nvm i figured it out. it was not in reference to n=4. equation used would be wavelength = 2L/n

Homework Statement

An electron is in an infinite potential well of width L. Which is not an allowed deBroglie wavelength for the electron to have when n=4?
wavelength(k) = 3L, 2L, L/2, or L/3

Homework Equations

k=h/p
would states in an infinite square well be useful too?

The Attempt at a Solution

The answer I was given is 3L which is throwing me off because I thought it would be L/3. there at 2 wavelengths at n=4. 3L= 6 wavelengths which is still integer. I was thinking L/3 because 2/3 is not a valid wavelength. Was the answer i was given wrong or am I doing something wrong?

 

[sankalpmittal]

nvm i figured it out. it was not in reference to n=4. equation used would be wavelength = 2L/n

Homework Statement

An electron is in an infinite potential well of width L. Which is not an allowed deBroglie wavelength for the electron to have when n=4?
wavelength(k) = 3L, 2L, L/2, or L/3

Homework Equations

k=h/p
would states in an infinite square well be useful too?

The Attempt at a Solution

The answer I was given is 3L which is throwing me off because I thought it would be L/3. there at 2 wavelengths at n=4. 3L= 6 wavelengths which is still integer. I was thinking L/3 because 2/3 is not a valid wavelength. Was the answer i was given wrong or am I doing something wrong?

Question is asking that "Which is not an allowed deBroglie wavelength for the electron to have when n=4?"

As you know that de-broglie wavelength is an integral multiple of h/mv.

Here you use formula ,
λ = 2L/n

If electron is excited , then there are infinite number of wavelengths.

If it is de-excited , then there will be only 6 combination of wavelengths ,
4-3,4-2, 4-1, 3-2, 3-1, 2-1.

Now the answer you were given was correct.

Hint : Let h/mv = R

Then ,

λ = R2L/n

2L/n must always be an Integer. In which option it will not be an integer ?