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Question: The electron energy in a hydrogen atom is given by \( {{{E}}_{{n}}} = ( - 2.18{{ }} \times {{ }}{10^...

The electron energy in a hydrogen atom is given by En=(−2.18×10−18)(1n2)joules{{{E}}_{{n}}} = ( - 2.18{{ }} \times {{ }}{10^{ - 18}})\left( {\dfrac{1}{{{n^2}}}} \right){{joules }} .Calculate the required energy to remove an electron completely from the n=2 orbit. What is the longest wavelength (in A) of light that can be used to cause this transition?

Explanation

Solution

In this question, we shall use the formula given to calculate the energy required to remove an electron from 2nd orbit and use Planck’s formula to convert it to frequency and then into wavelength. Wavelength of light is given as the ratio of the speed of light divided by the frequency of the light.

Formula Used:
ν=Eh{{\nu = }}\dfrac{{{E}}}{{{h}}}
λ=cν\lambda = \dfrac{{{c}}}{{{\nu }}}
Where E is the energy, c is the speed of light, λ\lambda is the wavelength and ν\nu is the frequency.

Complete step-by-step answer
In the question, the energy of electrons in a hydrogen atom is given by En=(−2.18×10−18)(1n2)joules{{{E}}_{{n}}} = ( - 2.18{{ }} \times {{ }}{10^{ - 18}})\left( {\dfrac{1}{{{n^2}}}} \right){{joules }} where n=2.
E2=(−2.18×10−18)(122){{{E}}_{{2}}} = ( - 2.18{{ }} \times {{ }}{10^{ - 18}})\left( {\dfrac{1}{{{2^2}}}} \right)
Therefore, the energy will be,
E2=−0.545×10−18joules{{{E}}_{{2}}} = - 0.545{{ }} \times {10^{ - 18}}{{ joules}} .
The negative sign indicates that the electron is in the bound state, so the energy of an electron in n=2 orbit is −0.545×10−18joules- 0.545{{ }} \times {10^{ - 18}}{{ joules}} . The energy required to remove electrons from n=2 orbit will be equal and opposite that it will be 0.545×10−18joules0.545{{ }} \times {10^{ - 18}}{{ joules}} .
The longest wavelength (in A) can be used to cause this transition, where energy is given by,
E=νh{{E = \nu h}} where,
E=Energy,ν=Frequency of light,h=Planck constant{{E = Energy, \nu = \text{Frequency of light}, h = \text{Planck constant}}}
Therefore,
ν=Eh{{\nu = }}\dfrac{{{E}}}{{{h}}}
Substituting the values in the equation,
ν=(0.545×10−18)J(6.626×10−34)Js\nu = \dfrac{{(0.545{{ }} \times {{ }}{{10}^{ - 18}}){{ J}}}}{{(6.626{{ }} \times {{ }}{{10}^{ - 34}}){{ Js}}}} ,
ν=8.2252×1014Hz\nu = 8.2252{{ }} \times {{ }}{10^{14}}{{ Hz}} .
Since,
ν=cλ,wherec=Speed of light and λ=Wavelength{{\nu = c}}\lambda {{, where c = \text{Speed of light and }}}\lambda {{ = \text{Wavelength} }} ,
Therefore,
λ=cν\lambda = \dfrac{{{c}}}{{{\nu }}} ,
Substituting the values in the equation,
λ=(3.0×108M/s)(8.2252×1014s−1)(Hz)\lambda = \dfrac{{(3.0 \times {{10}^8}{{ M/s}})}}{{(8.2252 \times {{10}^{14}}{{ }}{{{s}}^{{{ - 1}}}})({{Hz}})}} ,
λ=3.6473×10−7M\lambda = 3.6473{{ }} \times {{ }}{10^{ - 7}}{{ M}} .
Hence, the wavelength required in Angstroms will be,
λ=3.6473×10−7×1010Ao\lambda = 3.6473 \times {10^{ - 7}} \times {10^{10}}{{ }}{{{A}}^{{o}}} ,
i.e. λ=3647.3Ao{{\lambda = 3647}}{{.3 }}{{{A}}^{{o}}} .
Therefore, the longest wavelength (in A) of light that can be used to cause this transition will be λ=3647.3Ao\lambda = 3647.3{{ }}{{{A}}^{{o}}} .

Note:
The energy of an electron in n=2 orbit is −0.545×10−18joules- 0.545{{ }} \times {10^{ - 18}}{{ joules}} which is the potential energy of the electron under the influence of the nucleus. To remove the electron, equal energy is required which is +0.545×10−18joules+ 0.545{{ }} \times {10^{ - 18}}{{ joules}} , as the electron is removed completely from the nucleus the potential energy of electron becomes zero.