7. (a) Using the Bohr’s model, calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels.
Answer:
(a) Let v1 be the orbital speed of the electron in a hydrogen atom in the ground state level, n1 = 1. For charge (e) of an electron, v1 is given by the relation,
\(v_1 = \frac{e^{2}}{n_1 4\;\pi\;\epsilon_0\; (\frac{h}{2\pi}) } = \frac{e^{2}}{2\;\epsilon_0 \;h}\)
Where, e = 1.6 × 10−19 C
ϵ0 = Permittivity of free space = 8.85 × 10−12 N−1 C 2 m−2
h = Planck’s constant = 6.62 × 10−34 J s
\(v_1 = \frac{(1.6\times10^{-19})^{2}}{2\times8.85\times10^{-12}\times6.62\times10^{-34}}=\) 0.0218 x 108 = 2.18 × 106m/s
For level n2 = 2, we can write the relation for the corresponding orbital speed as:
\(v_2 = \frac{e^2}{n_2 2 \epsilon_0 h}= \frac{(1.6 \times 10^{-19})^2}{2 \times 2 \times 8.85 \times 10^{-12} \times 6.62 \times 10^{-34}}= \)1.09 × 106 m/s
For level n3= 3, we can write the relation for the corresponding orbital speed as:
\(v_3 = \frac{e^2}{n_3 2 \epsilon_0 h} = \frac{(1.6 \times 10^{-19})^2}{2 \times 3 \times 8.85 \times 10^{-12} \times 6.62 \times 10^{-34}}=\) 7.27 × 105 m/s
Therefore, in a hydrogen atom, the speed of the electron at different levels that is n = 1, n = 2, and n = 3 is 2.18 × 106m/s, 1.09×106m/s and 7.27 × 105 m/s.
(b) Let, T1 be the orbital period of the electron when it is in level n1= 1.
Orbital period is related to orbital speed as:
\(T_1 = \frac{ 2 \pi r_1 } { v_1 } \) [Where, r1 = Radius of the orbit] \(=\frac{ { n _ 1 } ^ 2 \;h ^ 2 \;\epsilon_0}{\pi \;m e^2 }\)
h = Planck’s constant = 6.62 × 10−34 J s
e = Charge on an electron = 1.6 × 10−19 C
ε0 = Permittivity of free space = 8.85 × 10−12 N−1 C2 m−2
m = Mass of an electron = 9.1 × 10−31 kg
\(T_1 = \frac{ 2 \pi r_1 } { v_1 } =\frac{2 \;\pi \;\times ( 1 ) ^ { 2 } \;\times \;( 6.62 \;\times \;10 ^ { -34 } ) ^ { 2 } \;\times \;8.85 \;\times \;10 ^{ -12 }}{ 2.18 \;\times \;10 ^ { 6 } \;\times \;\pi \;\times 9.1 \;\times \;10 ^ { -31 } \;\times \;( 1.6 \;\times \;10 ^ { -19 } ) ^ { 2 } }=\) 15.27 × 10-17 = 1.527 × 10-16 s
For level n2 = 2, we can write the period as:
\(T_2 = \frac{ 2 \pi r_2 } { v_2 } \) [Where, r2 = Radius of the electron in n2 = 2] \(=\frac{ {( n _ 2 )} ^ 2 \;h ^ 2 \;\epsilon_0}{\pi \;m e^2 }\)
\(=\frac{2 \;\pi \;\times \;( 2 ) ^ { 2 } \;\times \;( 6.62 \;\times \;10 ^ { -34 } ) ^ { 2 }\;\times \;8.85 \;\times 10 ^{ -12 }}{ 1.09 \;\times \;10 ^ { 6 } \;\times \;\pi \;\times \;9.1 \;\times \;10 ^ { -31 } \;\times \;( 1.6 \;\times \;10 ^ { -19 } ) ^ { 2 } }=\) 1.22 × 10-15s
For level n3 = 3, we can write the period as:
\(T_3= \frac{ 2 \pi r_3 } { v_3 }\) [Where, r3 = Radius of the electron in n3 = 3] \(=\frac{ {( n _ 3 )} ^ 2 \;h ^ 2 \;\epsilon_0}{\pi \;m e^2 }\)
\(=\frac{2 \;\pi \;\times \;( 3 ) ^ { 2 } \;\times \;( 6.62 \;\times \;10 ^ { -34 } ) ^ { 2 }\;\times \;8.85 \;\times 10 ^{ -12 }}{ 7.27 \;\times \;10 ^ { 5 } \;\times \;\pi \;\times \;9.1 \;\times \;10 ^ { -31 } \;\times \;( 1.6 \;\times \;10 ^ { -19 } ) ^ { 2 } }=\)= 4.12 × 10-15s
Therefore, 1.52 × 10-16s, 1.22 × 10-15s and 4.12 × 10-15s are the orbital periods in each levels.
8. The radius of the innermost electron orbit of a hydrogen atom is 5.3×10–11 m. What are the radii of the n = 2 and n = 3 orbits?
Answer:
The radius of the innermost orbit of a hydrogen atom, r1 = 5.3 × 10−11 m.
Let r2 be the radius of the orbit at n = 2. It is related to the radius of the inner most orbit as:
r2 = (n)2r1 = 4 × 5.3 × 10-11 = 2.12 × 10-10 m
For n = 3, we can write the corresponding electron radius as:
r3 = (n)2r1 = 9 × 5.3 x 10-11 = 4.77 × 10-10 m
Therefore, 2.12 × 10−10 m and 4.77 × 10−10 m are the radii of an electron for n = 2 and n = 3 orbits respectively.
9. A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
Answer:
It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is −13.6 eV.
When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes −13.6 + 12.5 eV i.e., −1.1 eV.
Orbital energy is related to orbit level (n) as:
\(E = \frac {-13.6 } { n ^ { 2 } } \; eV\)
For n = 3, E = -13.6 / 9 = – 1.5 eV
This energy is approximately equal to the energy of gaseous hydrogen.
It can be concluded that the electron has jumped from n = 1 to n = 3 level.
During its de-excitation, the electrons can jump from n = 3 to n = 1 directly, which forms a line of the Lyman series of the hydrogen spectrum.
We have the relation for wave number for Lyman series as:
\(\frac { 1 } { \lambda } = R_y \left ( \frac { 1 } { 1 ^ { 2 } } -\frac { 1 } { n ^ { 2 } } \right )\)
Where, Ry = Rydberg constant = 1.097 × 107 m−1
λ = Wavelength of radiation emitted by the transition of the electron
For n = 3, we can obtain λ as:
\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 1 } { 1 ^ { 2 } } -\frac { 1 } { 3 ^ { 2 } } \right )\)
\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( 1 -\frac { 1 } { 9 } \right )\)
\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 8 } { 9 } \right )\)
\(\lambda = \frac { 9 }{8 \times 1.097 \times 10 ^ {7 }}= 102.55 \,nm\)
If the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as:
\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 1 } { 1 ^ { 2 } } -\frac { 1 } { 2 ^ { 2 } } \right )\)
\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( 1 -\frac { 1 } { 4 } \right )\)
\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 3 } { 4 } \right )\)
\(\lambda = \frac { 4 }{3 \times 1.097 \times 10 ^ {7 }} = 121.54 \;nm\)
If the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is given as:
\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 1 } { 2 ^ { 2 } } -\frac { 1 } { 3 ^ { 2 } } \right )\)
\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 1 } { 4 } -\frac { 1 } { 9 } \right )\)
\(\frac { 1 } { \lambda } = 1.097 \times 10 ^ { 7 } \left ( \frac { 5 } { 36 } \right )\)
\(\lambda = \frac { 36 }{ 5 \times 1.097 \times 10 ^ {7 }}= 656.33 \,nm\)
This radiation corresponds to the Balmer series of the hydrogen spectrum.
Therefore, there are two wavelengths that are emitted in Lyman series and they are approximately 102.5 nm and 121.5 nm and one wavelength in the Balmer series which is 656.33 nm.