Find the energy of the ground state (n=1) of a proton in a one-dimensional box o

Question

Find the energy of the ground state (n=1) of a proton in a one-dimensional box of length 5.2 nm. Answer in units of meV. Calculate the wavelength of electromagnetic radiation emitted when the photon makes a transition from n=2 to n=1. Answer in units of mu m. Calculate the wavelength of electromagnetic radiation emitted when the photon makes a transition from n=3 to n=2. Answer in units of mu m. Calculate the wavelength of electromagnetic radiation emitted when the photon makes a transition from n=3 to n=1. Answer in units of mu m.

Explanation / Answer

1) Given: proton is in ground state (n=1) in a one dimensional box of length 5.2 nm.

E = (n²²²)/(2ma²)
n=1, is planck's reduced constant, = h/2 = 1.0546 × 10-27 cm2 g s-1

m is mass of particle ( mass of proton = 1.672 x 10^-27),

a is length of box = 5.2 nm.

Energy in ground state E_g in meV= (1²²²)/(2*1.672x 10^-27 * (5.2 x 10^-9)²)*(1.6 x 10^-19) = 758721.4123 meV

2-4) Use the above equation to figure out the differences in energy like :
(n'²-n²)(²²)/(2ma²)
where n' is initial, and n is final. Then use the equation
E = hf = hc/

wavelength = hc/E
where h is plancks constant, c is the speed of light 3*10^8 m/s, is wavelength, and f is frequency.

for n = 2 to n=1 ; E = (2^2 - 1)*E_g = 3*758721.4123 meV = 2276164.237 meV

,wavelength = 6.61 x10^-27 * 3*10^8 / E = .016335 um

for n = 3 to n=2 ; E3 = (3^2 - 4)*E_g = 5*758721.4123 meV

,wavelength = 6.61 x10^-27 * 3*10^8 / E3 = 0.003267 um

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