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Question: What is the energy in joules required to shift the electron of the hydrogen atom from the first Bohr...

What is the energy in joules required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is −2.18×10−11ergs-2.18\times {{10}^{-11}}ergs.

Explanation

Solution

We can calculate the energy for the hydrogen electron by applying the energy formula of Bohr as:  En=−2.18×10−11n2ergs\ {{E}_{n}}=-\dfrac{2.18\times {{10}^{-11}}}{{{n}^{2}}}ergs and ΔE=E5−E1\Delta E={{E}_{5}}-{{E}_{1}} and if we know, the energy then we can easily find the wavelength by applying the formula as: λ=hcE \lambda =\dfrac{hc}{E}\text{ }. Now solve it.

Complete step by step solution:
This numerical is based on the energy of the electron given by Bohr. According to the Bohr, the electrons revolve around the nucleus in that fixed orbit and has a definite value of energy and these orbits are also known as energy levels and the energy of the electrons in that particular orbit is fixed and doesn’t change with time. The different energy levels are numbered as 1,2,3,4,5---etc. starting from the nucleus. The energy of each orbit is given by the expression as:
 En=−2π2me4n2h2\ {{E}_{n}}=-\dfrac{2{{\pi }^{2}}m{{e}^{4}}}{{{n}^{2}}{{h}^{2}}}
Substituting the values, the values of m (mass of the electron), e (charge on the electron) and h (Planck’s constant), we get:
 En=−21.8×10−19n2J/atom\ {{E}_{n}}=-\dfrac{21.8\times {{10}^{-19}}}{{{n}^{2}}}J/atom
⇒ −1312n2kJmole−1\Rightarrow \text{ }-\dfrac{1312}{{{n}^{2}}}kJmol{{e}^{-1}}
⇒ −13.6n2eV/atom (1eV=1.602 ×J)\Rightarrow \text{ }-\dfrac{13.6}{{{n}^{2}}}eV/atom\text{ }(1eV=1.602\text{ }\times J)
Where n=1,2,3, ----etc. stands for the 1st,2 Nd, 3rd ----etc. levels. The energy level which is closest to the nucleus has the lowest energy and the energy increases as the energy levels increase and so on.
But for hydrogen like particles, the expression for energy is:
 En=−2π2mZ2e4n2h2 ⇒ −1312Z2n2kJmole−1 ⇒ −2.18×10−11n2ergs \begin{aligned} & \ {{E}_{n}}=-\dfrac{2{{\pi }^{2}}m{{Z}^{2}}{{e}^{4}}}{{{n}^{2}}{{h}^{2}}} \\\ & \Rightarrow \text{ }-\dfrac{1312{{Z}^{2}}}{{{n}^{2}}}kJmol{{e}^{-1}} \\\ & \Rightarrow \text{ }-\dfrac{2.18\times {{10}^{-11}}}{{{n}^{2}}}ergs \\\ \end{aligned}
Now considering the statement;
We can find the energy of the hydrogen electron by applying the formula as;
 En=−2.18×10−11n2ergs\ {{E}_{n}}=-\dfrac{2.18\times {{10}^{-11}}}{{{n}^{2}}}ergs
Energy of the hydrogen atom in first Bohr orbit E1=−2.18×10−1112=−2.18×10−111ergs\ {{E}_{1}}=-\dfrac{2.18\times {{10}^{-11}}}{{{1}^{2}}}=-\dfrac{2.18\times {{10}^{-11}}}{1}ergs
Energy of the hydrogen atom in fifth Bohr orbit E5=−2.18×10−1152=−2.18×10−1125ergs\ {{E}_{5}}=-\dfrac{2.18\times {{10}^{-11}}}{{{5}^{2}}}=-\dfrac{2.18\times {{10}^{-11}}}{25}ergs
Energy required to shift the electron of the hydrogen atom from first to fifth Bohr orbit is as;
 ΔE=E5−E1 ⇒ −2.18×10−1125−2.18×10−111 ⇒ 2.18×10−11(11−125) ⇒ 2.18×10−11(25−125) ⇒ 2.18×10−11(2425) ⇒ 2.09×10−11 ergs ⇒ 2.09×10−18J (1 erg=107J) \begin{aligned} & \ \Delta E={{E}_{5}}-{{E}_{1}} \\\ & \Rightarrow \text{ }-\dfrac{2.18\times {{10}^{-11}}}{25}-\dfrac{2.18\times {{10}^{-11}}}{1} \\\ & \Rightarrow \text{ }2.18\times {{10}^{-11}}\left( \dfrac{1}{1}-\dfrac{1}{25} \right) \\\ & \Rightarrow \text{ }2.18\times {{10}^{-11}}\left( \dfrac{25-1}{25} \right) \\\ & \Rightarrow \text{ }2.18\times {{10}^{-11}}\left( \dfrac{24}{25} \right) \\\ & \Rightarrow \text{ }2.09\times {{10}^{-11}}\text{ }ergs \\\ & \Rightarrow \text{ }2.09\times {{10}^{-18}}J\text{ }(1\text{ }erg={{10}^{7}}J) \\\ \end{aligned}
So, the energy in joules required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit is 2.09×10−18J2.09\times {{10}^{-18}}J.
Now, calculating the wavelength of the light emitted as;
As we know;
 E=hv ⇒ hcλ (λ=cv)  λ=hcE ————–(1) \begin{aligned} & \ E=hv \\\ & \Rightarrow \text{ }h\dfrac{c}{\lambda }\text{ (}\lambda \text{=}\dfrac{c}{v}) \\\ & \ \lambda =\dfrac{hc}{E}\text{ --------------(1)} \\\ \end{aligned}
So, as we know that;
 E=2.09×10−11 ergs\ E=2.09\times {{10}^{-11}}\text{ }ergs
 h=6.62×10−27 erg sec⁡−1\ h=6.62\times {{10}^{-27}}\text{ }erg\text{ }{{\sec }^{-1}}
 c=3×1010 cm sec⁡−1\ c=3\times {{10}^{10}}\text{ cm }{{\sec }^{-1}}
then, put all these values in equation (1), we get;
 λ=6.62×10−27 erg sec⁡−1 × 3×1010 cm sec⁡−12.09×10−11 ergs ⇒ 9.51×10−6 cm \begin{aligned} & \ \lambda =\dfrac{6.62\times {{10}^{-27}}\text{ }erg\text{ }{{\sec }^{-1}}\text{ }\times \text{ }3\times {{10}^{10}}\text{ cm }{{\sec }^{-1}}}{2.09\times {{10}^{-11}}\text{ }ergs} \\\ & \Rightarrow \text{ 9}\text{.51}\times {{10}^{-6}}\text{ cm} \\\ \end{aligned}

Hence, the wavelength of the light emitted when the electron returns to the ground state is 9.51×10−6 cm\text{9}\text{.51}\times {{10}^{-6}}\text{ cm}.

Note: Don’t forget to convert the units i.e. ergs into the joules as the energy of the electron of the hydrogen atom is asked in joules and always convert the units accordingly as mentioned in the given statement.