Physics: An electron is confined in a 3.0 nm long one dimensional box. the electron in this energy state has a wavelength of 1.0 nm. a) what is the quantum number of this electron? b) what is the ground state energy of this electron in a box c) what is the photon wavelength that is emitted in a transition from the energy level in part a to the first excited state?

An electron is confined in a 3.0 nm long one dimensional

Ответ:

brownvester44

01.01.2022

Physics

Explanation:

It is given that,

Length of one dimensional box, $l=3\ nm=3\times 10^{-9}\ m$

Wavelength of electron, $\lambda=1\ nm=10^{-9}\ m$

(a) We need to find the quantum number of this electron. The energy of electron in one dimensional box is given by :

$E=\dfrac{n^2h^2}{8ml^2}$

Also, energy is given by, $E=\dfrac{hc}{\lambda}$

$\dfrac{hc}{\lambda}=\dfrac{n^2h^2}{8ml^2}$

$n^2=\dfrac{8cml^2}{h\lambda}$

c is the speed of light

m is the mass of electron

h is the Planck's constant

$n^2=\dfrac{8\times 3\times 10^8\times 9.1\times 10^{-31}\times (3\times 10^{-9})^2}{6.63\times 10^{-34}\times 10^{-9}}$

$n=\sqrt{29647.05}$

n = 172.18

n = 172

(b) For ground state energy, n = 1

$E=\dfrac{(6.64\times 10^{-34})^2}{8\times 9.1\times 10^{-31}\times (3\times 10^{-9})^2}$

$E=6.72\times 10^{-21}\ J$

(c) We need to find the wavelength of a photon that is emitted in a transition from the energy level in part a to the first excited state (n=2). So,

$\dfrac{hc}{\lambda}=E_{172}-E_2$

$\dfrac{hc}{\lambda}=\dfrac{(172)^2h^2}{8ma^2}-\dfrac{4h^2}{8ma^2}$

$\dfrac{hc}{\lambda}=\dfrac{h^2}{8ml^2}(29647.05-4)$

$\dfrac{hc}{\lambda}=\dfrac{h^2}{8ml^2}\times 29643.05$

$\lambda=\dfrac{8ml^2c}{h\times 29643.05}$

$\lambda=\dfrac{8\times 9.1\times 10^{-31}\times (3\times 10^{-9})^2\times 3\times 10^8}{6.63\times 10^{-34}\times 29643.05}$

$\lambda=1.0001\times 10^{-9}\ m$

Hence, this is the required solution.

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Ответ:

brentking2001

07.08.2021

Physics

The distance is 58.03 m

Explanation:

Constant Acceleration Motion

It occurs when the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

v_f=v_o+at

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

The conditions of the problem state the cyclist has an initial speed of v0=20.7 m/s during t=7.8 seconds and acceleration of -3.4 m/s^2.

The final speed is:

$v_f=20.7+(-3.4)\cdot 7.8$

v_f=20.7-26.52

$v_f=-5.82\ m/s$

Note the cyclist has stopped and come back because his speed is negative. Now calculate the distance:

$\displaystyle x=20.7\cdot 7.8+\frac{(-3.4)\cdot 7.8^2}{2}$

$\displaystyle x=161.46-103.43$

x=58.03 m

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