The energy of the electron in the lowest level of the hydrogen atom (n=1) is -2.

Question

The energy of the electron in the lowest level of the hydrogen atom (n=1) is -2.179×10-18 J. What is the energy of the electron in level n=4?

The electron in a hydrogen atom moves from level n=5 to level n=2. emitted: Is a photon emitted or absorbed? What is the wavelength of the photon?

An excited hydrogen atom emits a photon with a wavelength of 93.8 nm. ultraviolet: In what region of the spectrum is this photon? What is the frequency of this photon?

A hydrogen atom in the ground state absorbs a photon of wavelength 93.8 nm. What energy level does the electron reach? This excited atom then emits a photon of wavelength 1094.0 nm. What energy level does the electron fall to?

Explanation / Answer

1. In the hydrogen atom, with Z = 1, the energy of the emitted photon can be found using:

E = -13.6 eV *1/n^2

For n=1

E1 = -13.6 eV *1/n = -13.6 eV *1/1 = -13.6 eV = -2.179*10^-18 J

For n= 4

E1 = -13.6 eV *1/n = -13.6 eV *1/4 = -13.6 eV/4^2 = - 0.85J

2. When the electron in a hydrogen atom moves from level n=5 to level n=2. Photon is emitted

Energy of photon , E = -13.6 eV *(1/nf^2 - 1/ni^2) = -13.6 eV *(1/5^2 - 1/2^2) = 2.856 eV = 4.58*10^-19 J

Wavelength , = hc/E = (6.626 x 10 – 34*3*10^8)/( 4.58*10^-19) = 4.34*10^-7 m = 434 nm

3. Ultraviolet region falls in wavelength range 400 nm to 1 nm. Thus wavelength 93,8nm falls in this region..

f= c/ = (3*10^8)/(93.8*10^-9) = 3.2*10^15 Hz

4. When a hydrogen atom in the ground state absorbs a photon of wavelength 93.8 nm

Its energy E = hf = (6.626 x 10 – 34*3.2*10^15) = 2.12*10^-18 J = 13.23 eV

Now use equation,

E = -13.6 eV *(1/nf^2 - 1/ni^2)

Put n=1 for ground level

13.23 eV = -13.6 eV *(1/nf^2 - 1/1^2)    => n = 6

Now transition from n= 6 to n=...?

E = -13.6 eV *(1/nf^2 - 1/ni^2)

hc/ = -13.6 eV *(1/nf^2 - 1/ni^2)

(6.626 x 10 – 34*3 *10^8) /(1094*10^-9)= -2.179*10^-18 *(1/nf^2 - 1/6^2)      => nf = 4

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