Question

What is the quantum number of an electron confined in a 3.0-nm-long one-dimensional box if the electron's de broglie wavelength is 1.0 nm ? n =?

Answer 1

Answer:

n = 6

Explanation:

To find the quantum number you use the following formula:

$E_n=\frac{n^2\pi^2\hbar^2}{2m_eL^2}$    (1)

n: quantum number

En: energy of the nth state

L: length of the box

hbar = 1,05*10^-34Js

me: mass of the electron = 9.1*10^{-31}kg

Thus, you need to calculate En. To find En you use the wavelength of the electron:

$p=\frac{h}{\lambda}\\\\E_n=\frac{p^2}{2m_e}=\frac{h^2}{2\lambda^2 m_e}$

$E_n=\frac{(6.62*10^{-34}Js)^2}{2(1.0*10^{-9}m)^2(9.1*10^{-31}kg)}=2.40*10^{-19}J$

By replacing En and doing n the subject of the formula (1) you obtain:

$n=\sqrt{\frac{2m_eL^2E_n}{\pi^2 \hbar^2}}=\sqrt{\frac{2(9.1*10^{-31}kg)(3*10^{-9}m)^2(2.40*10^{-19})}{\pi^2(1.05*10^{-34}Js)^2}}=6$

hence, the quantum number is 6

Answer 2

The quantum number of the electron confined in the given box is 6.

The given parameters;

• wavelength, λ = 1 nm
• length of the box, L = 3 nm

The energy of the electron is calculated as shown below;

$E_n = \frac{P^2}{2m_e} = \frac{h^2}{2\lambda ^2 m_e} = \frac{(6.626\times 10^{-34})^2}{2 \times (1\times 10^{-9})^2 \times (9.11\times 10^{-31})} \\\\E_n = 2.41 \times 10^{-19} \ J$

The quantum number of the electron is calculated as follows;

$E_n = \frac{n^2 \bar h^2 \pi^2}{2m_e L^2} \\\\n^2 = \frac{2m_e L^2 E_n}{\bar h^2 \pi^2 } \\\\n = \sqrt{\frac{2m_e L^2 E_n}{\bar h^2 \pi^2 } } \\\\n = \sqrt{\frac{2\times 9.11 \times 10^{-31} \times (3\times 10^{-9})^2 \times (2.41\times 10^{-19})}{(1.05 \times 10^{-34})^2 \times (\pi)^2} } \\\\n = 6$

Thus, the quantum number of the electron confined in the given box is 6.

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