contestada

When the electron in a hydrogen atom moves from n = 6 to n = 1, light with a wavelength of ________ nm is emitted?

Respuesta :

SebRydahl
First we find the energy level with the following formula, where a is the energy level, n1 is the final energy level, n2 is the starting energy level and r is Rydberg's constant in Joules
[tex]a = r \times ( \frac{1}{n1} - \frac{1}{n2} )[/tex]
We insert the values

[tex]a = 2.18 \times {10}^{ - 18} \times ( \frac{1}{ {1}^{2} } - \frac{1}{ {6}^{2} } )[/tex]
[tex] = 2.12 \times {10}^{ - 18} [/tex]
The wavelength is found with this formula, where h is Planck's constant and c is the speed of light
[tex]wavelength = \frac{h \times c}{a} [/tex]
Finally we insert the values
[tex] \frac{6.626 \times {10}^{ - 34} \times 3 \times {10}^{8} }{2.12 \times {10}^{ - 18} } = 9.376 \times {10}^{ - 8} [/tex]
Which is the same as 93.8 nm

When the electron in a hydrogen atom moves from n = 6 to n = 1, light with a wavelength of 93.8nm is emitted.

Explanation:

Given :

n = 6 and n = 1

Calculation :

Wavelength can be find by using below formula

[tex]\rm Wavelength = \dfrac{h\times c}{a}[/tex] ---- (1)

Whare,

h is the plank constant = [tex]6.626\times 10^-^3^4[/tex]

c is the speed of light = [tex]3 \times 10^8[/tex]

and a is the energy level,

[tex]a = r\times (\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})[/tex]

where,

rydberg's constant, [tex]\rm r = 2.18\times10^-^1^8[/tex]

[tex]a = 2.18\times10^-^1^8(\dfrac{1}{1^2}-\dfrac{1}{6^2})[/tex]

[tex]a =2.12\times 10^-^1^8[/tex]

From equation (1),

[tex]\rm Wavelength = \dfrac{6.626\times10^-^3^4\times 3\times 10^8}{2.12\times 10^-^1^8}=9.376\times10^-^8[/tex]

When the electron in a hydrogen atom moves from n = 6 to n = 1, light with a wavelength of 93.8nm is emitted.

For more information, refer the link given below

https://brainly.com/question/13533093?referrer=searchResults

Otras preguntas