Question

What is the deBroglie wavelength associated with a particle in a 1D box with
width of 5 nm, in its first excited state?​

Answer 1

Answer:

10nm

Explanation:

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Answer 2

The deBroglie wavelength associated with a particle in a 1D box is 5nm

Explanation:

Given that,

A particle is in a one dimensional box with a width of 5nm and it is in first excited state

Let width of box=W=5nm

The energy of particle in one dimensional box is given as

$E = \frac{n {}^{2} h {}^{2} }{8mW {}^{2} }$

where,

n=excited state orbit number

h=planck's constant

m=mass of particle

Also energy can be related to momentum as

$E = \frac{p {}^{2} }{2m } \\ \frac{p {}^{2} }{2m } = \frac{n {}^{2} h {}^{2} }{8mW {}^{2} } \\ = > W {}^{2} = \frac{n {}^{2}h {}^{2} }{4p { }^{2} }$

For first excited state,n=2

$= > W {}^{2} = \frac{h {}^{2} }{p {}^{2} } = > W = \frac{h}{p}$

we know that de-broglie wavelength is

$λ = \frac{h}{p } \\ = > λ = W = 5nm$

Therefore,deBroglie wavelength associated with a particle in a 1D box is 5nm

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