Question

1st excited state energy of a particle
in an infinite height potential well is 16
MeV. If width of the potential well is
doubled, first excited state energy
will become​

Answer 1

Given that,

First excited state energy of particle = 16 MeV

Let the width of the potential well is  L.

We know that,

The excited state energy is

$E_{n}=\dfrac{h^2}{8mL^2}$

Where, h = planck's constant

m = mass of particle

L = width of well

If width of the potential well is  doubled,

We need to calculate the first excited state energy

Using formula of excited state energy  

$E=\dfrac{h^2}{8mL^2}$

for both width,

$\dfrac{E_{1}}{E_{2}}=\dfrac{\dfrac{h^2}{8mL^2}}{\dfrac{h^2}{8m(2L)^2}}$

Put the value into the formula

$\dfrac{E_{1}}{E_{2}}=\dfrac{h^2}{8mL^2}\times\dfrac{8m(2L)^2}{h^2}$

$\dfrac{16}{E_{2}}=\dfrac{4}{1}$

$E_{2}=4\ MeV$

Hence, first excited state energy  will become​ 4 MeV.

Answer 2

Answer:

1st excited-state energy of a particle  in an infinite height potential well is 16

MeV. If the width of the potential well is  doubled, first excited state energy

will become​ $4MeV$

Explanation:

• The nth state energy of a particle in an infinite potential well of width L is given by the formula

       $E_{n} =\frac{n^{2} h^{2} }{8mL^{2} }$

       where $n$-state of the particle

       $m$-the mass of the particle

       $h$-Planck's constant

      $L$-width of the potential well

• that is energy is inversely proportional to the square of the width

From the question, it is given that

the energy of the particle in the first excited state ($n=2$) is

$E_{2} =\frac{4h^{2} }{8mL^{2} }=16MeV$

then the width of the potential well is doubled

⇒$L_{new}=2L$

so the new energy will be

$E_{new} =\frac{4 h^{2} }{8m(2L)^{2} }=\frac{1}{4} \frac{4h^{2} }{8mL^{2} }$

⇒$E_{new}=E_{2} /4$

⇒           $=16/4=4MeV$