Question

Given that RH= 2.18 x 10⁻¹⁸J, 1 nm = 1 x 10⁻⁹m, h = 6.63 x 10⁻³⁴J·s, and c = 3.00 x 10⁸m/s:
Calculate the wavelength, innanometers, of the light emitted by a hydrogen atom during a transition of its electron from the n = 7 to the n = 3 principal energy level.

Answer 1

Answer:

The wavelength the light emitted by a hydrogen atom during a transition is 1006 nm.

Explanation:

By using Rydberg's Equation we cab determine the wavelength of the light:

$\Delta E=R_H\times Z^2\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\Delta E$ = Energy difference

$R_H$ = Rydberg's Constant

$n_f$ = Final energy level

$n_i$= Initial energy level

We have : $n_i=7,n_f=3$ , Z = 1

$R_H=2.18\times 10^{-18} J$

$\Delta E=2.18\times 10^{-18} J\times 1^2\left(\frac{1}{7^2}-\frac{1}{3^2} \right )$

$\Delta E=1.9773\times 10^{-19} J$

Now by using Plank's equation we can determine the wavelength of the light emitted.

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

E = Energy of the emitted light

h = Planck's constant = $6.63\times 10^{-34} Js$

c = speed of light = $3.00\times 0^8 m/s$

For the given transition the energy of the light = E

$E =1.9773\times 10^{-19} J$

$\lambda=\frac{hc}{E}=\frac{6.63\times 10^{-34} Js\times 3.00\times 0^8 m/s}{1.9773\times 10^{-19} J}$

$\lambda =1.006\times 10^{-6} m =1.006\times 10^{-6}\times 10^9=1006 nm$

The wavelength the light emitted by a hydrogen atom during a transition is 1006 nm.