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Calculate the wavelength of the electromagnetic radiation required to excite an electron from the ground state to the level with

Novosadov [1.4K]

The question is incomplete. The complete question is :

Calculate the wavelength of the electromagnetic radiation required to excite an electron from the ground state to the level with n = 6 in a one-dimensional box 34.0 pm in length.

Solution :

In an one dimensional box, energy of a particle is given by :

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

Here, h = Planck's constant

n = level of energy

= 6

m = mass of particle

a = box length

For n = 6, the energy associated is :

$\Delta E = E_6 - E_1$

$\Delta E = \left( \frac{n_6^2h_2}{8ma^2}\right) - \left( \frac{n_1^2h_2}{8ma^2}\right)$

$=\frac{h^2(n_6^2 - n_1^2)}{8ma^2}$

We know that,

$E = \frac{hc}{\lambda}$

Here, λ = wavelength

h = Plank's constant

c = velocity of light

So the wavelength,

$= \frac{hc}{E}$

$=\frac{hc}{\frac{h^2(n_6^2 - n_1^2)}{8ma^2}}$

$=\frac{8ma^2c}{h(n_6^2 - n_1^2)}$

$=\frac{8 \times 9.109 \times 10^{-31}(0.34 \times 10^{-10})^2 (3 \times 10^8)}{6.626 \times 10^{-34} \times (36-1)}$

$= \frac{ 8 \times 9.109 \times 0.34 \times 0.34 \times 3 \times 10^{-43}}{6.626 \times 35 \times 10^{-34}}$

$=\frac{25.27 \times 10^{-43}}{231.91 \times 10^{-34}}$

$= 0.108 \times 10^{-9}$ m

= 108 pm

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