Calculate the wavelength, in nanometers, of the light emitted by a hydrogen atom when its electron falls from the n = 7 to the n = 4 principal energy level. Recall that the energy levels of the H atom are given by En = –2.18 × 10–18 J(1/n2)

Answers

Answer 1

Answer:

Answer:

The wavelength of the light emitted by electron fall will be 216.62 nm.

Explanation:

The change in the energy state of the matter can be obtained from the principal quantum number of the shells.

Energy of 4th shell - Energy of 7th shell = $\rm \Delta$ E

$\Delta$ E = $\rm 2.18\;*\;10^-^1^8\;J\;\left ( (1)/(n_f^2) \;-\;(1)/(n_i^2) \right )$

$\rm \Delta\;E\;=\;-2.18\;*\;10^-^1^8\;\left ((1)/(4^2) \;-\;(1)/(7^2) \right )$

$\rm \Delta\;E\;=\;-2.18\;*\;10^-^1^8\;*\;0.0421$

$\rm \Delta\;E\;=\;0.0917\;*\;10^-^1^8$ J

The wavelength can be calculated as:

$\rm \Delta\;E\;=\;(hc)/(\lambda)$

where, h is Plank's constant,

c is speed of light

$\lambda$ is the wavelength

$\rm 0.0917\;*\;10^-^1^8\;J\;=\;(6.626\;*\;10^-^3^4\;J.s\;*\;2.998\;*\;10^8\;m/s)/(\lambda)$

$\rm\lambda\;=\;216.62nm$ .

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Answer 2

Answer:

Answer:

The wavelength of the light emitted by a hydrogen atom for the given transition is 2166 nm.

Explanation:

The energy of nth energy levels of the H atom is given as:

$E_n = -2.18 * 10^(-18) * (1)/(n^2) J$

Energy of the seventh energy level = $E_7$

$E_7=-2.18 * 10^(-18) * (1)/(7^2) J$

$E_7=-2.18 * 10^(-18) * (1)/(7^2) J=-4.4490* 10^(-20) J$

Energy of the seventh energy level = $E_4$

$E_4=-2.18 * 10^(-18) * (1)/(4^2) J$

$E_4=-2.18 * 10^(-18) * (1)/(16) J=-1.3625* 10^(-19) J$

Energy of the light emitted will be equal to the energy difference of the both levels.

$E=E_7-E_4=-4.4490* 10^(-20) J-(-1.3625* 10^(-19) J)$

$E=9.176* 10^(-20) J$

Wavelength corresponding to energy E can be calculated by using Planck's equation:

$E=(hc)/(\lambda )$

$\lambda =(hc)/(E)=(6.626* 10^(-34) Js* 3* 10^8 m/s)/(9.176* 10^(-20) J)=2.166* 10^(-6) m=2166 nm$

The wavelength of the light emitted by a hydrogen atom for the given transition is 2166 nm.

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