Question

An electron is trapped in a one-dimensional infinite well of width 340 pm and is in its ground state. What are the (a) longest, (b) second longest, and (c) third longest wavelengths of light that can excite the electron from the ground state via a single photon absorption?

Answer 1

Answer:

(a) 1.2703×10⁻⁷ m

(b) 4.7636×10⁻⁸ m

(c) 2.5406×10⁻⁸ m

Explanation:

Given:

Width of the infinite well, L = 340 pm = 340×10⁻¹² m.

The formula for energy of the electron in nth state is:

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

The expression for the difference in energy between the levels having quantum numbers n(initial) to n(final) is:

$\Delta E_n=\frac {({n_f}^2-{n_i}^2)\times h^2}{8mL^2}$

According to Planks theory:

E = hv

where, v is the frequency

Also,

Frequency×Wavelength = Speed of light

So,

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

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

Also,  using energy from above formula as:

$\lambda=\frac {hc}{\frac {({n_f}^2-{n_i}^2)\times h^2}{8mL^2}}$

$\lambda=\frac {c\times {8mL^2}} {({n_f}^2-{n_i}^2)\times h}}$

For longest wavelength ni = 1 and nf = 2

m= mass of the electron = 9.1 ×10⁻³¹kg

c = 3×10⁸m/s

h = 6.625×10⁻³⁴J.sec

$\lambda_{Longest}=\frac {3\times 10^8\times {8\times 9.1\times 10^{-31}(340\times 10^{-12})^2}} {({2}^2-{1}^2)\times 6.625\times 10^{-34}}}$

Longest wavelength = 1.2703×10⁻⁷ m

For second longest wavelength ni = 1 and nf = 3

$\lambda_{Second\ Longest}=\frac {3\times 10^8\times {8\times 9.1\times 10^{-31}(340\times 10^{-12})^2}} {({3}^2-{1}^2)\times 6.625\times 10^{-34}}}$

Second longest wavelength = 4.7636×10⁻⁸ m

For third longest wavelength ni = 1 and nf = 4

$\lambda_{Third\ Longest}=\frac {3\times 10^8\times {8\times 9.1\times 10^{-31}(340\times 10^{-12})^2}} {({4}^2-{1}^2)\times 6.625\times 10^{-34}}}$

Third longest wavelength = 2.5406×10⁻⁸ m