A hydrogen atom is in its third excited state (n=4). Using Bohr theory of the atom calculate (a) the radius of the orbit, (b) the linear momentum of the electron, (c) the angular momentum of the electron (d) the kinetic energy (e) the potential energy, and (f) the total energy. (g) if the electron jumps to the ground state, what is the wavelength of the emitted photon?
To solve this problem using the Bohr theory of the atom, we need to utilize the equations and concepts related to the quantization of energy levels in an atom.
(a) Radius of the orbit:
According to Bohr's theory, the radius of the orbit of an electron in the nth energy level of a hydrogen atom is given by the formula:
r = (0.529 Å) * n²
where n is the principle quantum number.
For the given case, the hydrogen atom is in its third excited state (n=4), so we can substitute n=4 into the formula:
r = (0.529 Å) * (4²)
r = 2.116 Å
Therefore, the radius of the orbit is 2.116 Å.
(b) Linear momentum of the electron:
The linear momentum of an electron can be calculated using the formula:
p = m * v
where p is the momentum, m is the mass of the electron, and v is the velocity of the electron.
In Bohr's theory, the velocity of the electron in the nth energy level is given by:
v = (2.19 * 10⁶ m/s) / n
Substituting n=4 into the formula:
v = (2.19 * 10⁶ m/s) / 4
v = 5.475 * 10⁵ m/s
Now, we can calculate the linear momentum:
p = (9.10938356 x 10⁻³¹ kg) * (5.475 * 10⁵ m/s)
p ≈ 4.988 * 10⁻²⁶ kg·m/s
Therefore, the linear momentum of the electron is approximately 4.988 * 10⁻²⁶ kg·m/s.
(c) Angular momentum of the electron:
The angular momentum of an electron in the nth energy level can be calculated using the formula:
L = n * h / (2π)
where L is the angular momentum, n is the principle quantum number, and h is the Planck's constant.
Substituting n=4 and using the value of h as 6.62607015 x 10⁻³⁴ J·s:
L = (4 * 6.62607015 x 10⁻³⁴ J·s) / (2π)
L ≈ 4.195 * 10⁻³³ J·s
Therefore, the angular momentum of the electron is approximately 4.195 * 10⁻³³ J·s.
(d) Kinetic energy:
The kinetic energy of the electron in the nth energy level can be calculated using the formula:
K.E. = (-(2.18 * 10⁻¹⁸ J) / n²)
Substituting n=4 into the formula:
K.E. = (-(2.18 * 10⁻¹⁸ J) / (4²))
K.E. ≈ -3.0425 * 10⁻¹⁹ J
Therefore, the kinetic energy of the electron is approximately -3.0425 * 10⁻¹⁹ J.
(e) Potential energy:
The potential energy of an electron in the nth energy level of a hydrogen atom can be calculated using the formula:
P.E. = -(2.18 * 10⁻¹⁸ J) / n
Substituting n=4 into the formula:
P.E. = -(2.18 * 10⁻¹⁸ J) / 4
P.E. ≈ -5.45 * 10⁻¹⁹ J
Therefore, the potential energy of the electron is approximately -5.45 * 10⁻¹⁹ J.
(f) Total energy:
The total energy of the electron in the nth energy level of a hydrogen atom is the sum of its kinetic and potential energy:
Total energy = K.E. + P.E.
Substituting the calculated values:
Total energy = (-3.0425 * 10⁻¹⁹ J) + (-5.45 * 10⁻¹⁹ J)
Total energy ≈ -8.4925 * 10⁻¹⁹ J
Therefore, the total energy of the electron is approximately -8.4925 * 10⁻¹⁹ J.
(g) Wavelength of the emitted photon:
When an electron jumps from a higher energy level to a lower energy level, it emits a photon with a specific wavelength. The wavelength can be calculated using the formula:
λ = c / ν
where λ is the wavelength, c is the speed of light (approximately 3.00 x 10⁸ m/s), and ν is the frequency.
To find the frequency, we can use the energy difference between the energy levels:
ΔE = E_final - E_initial
The energy difference between the third excited state (n=4) and the ground state (n=1) is:
ΔE = -(2.18 * 10⁻¹⁸ J) * ((1/1²) - (1/4²))
Simplifying the expression:
ΔE = -(1.91 * 10⁻¹⁸ J)
Using the relation ΔE = h * ν, we can find the frequency:
ν = ΔE / h
ν = (-(1.91 * 10⁻¹⁸ J)) / (6.62607015 x 10⁻³⁴ J·s)
ν ≈ -2.888 x 10¹⁵ Hz
Now, we can calculate the wavelength:
λ = (3.00 x 10⁸ m/s) / (-2.888 x 10¹⁵ Hz)
λ ≈ -1.038 x 10⁻⁷ m
Since wavelength cannot be negative, we take the absolute value:
λ ≈ 1.038 x 10⁻⁷ m
Therefore, the wavelength of the emitted photon is approximately 1.038 x 10⁻⁷ meters.
r(n) = ε(o) •h^2•n^2/π•m•e^2 =
=5.3•10^-11•n^2 =5.3•10^-11•16 =
=8.48•10^-10 m.
v(n) = e^2/2• ε(o) •h•n=
=2.19•10^6/n =5.475•10^5 m/s.
p(n)=e•v(n) =1.6•10^-19•5.475•10^5 =
=8.77•106-14 kg•m/s.
L =m•v(n)•r(n) =n•h/2• π =4.22•10^-34 kg•m^2/s.
E(total) = -(1/4•π•ε(o)) •(e^2/2•r(n)) = -13.6/n^2 =
= - 0.85 eV = - 13.6•10^-19 J.
KE = - E(total) = = + 0.85 eV = + 13.6•10^-19 J.
PE = 2•E(total) = - 1.7 eV = - 2.72•10^-19 J.
1/λ = R•(1/1^2 – 1/n^2) =3•R/4,
λ =4/3•R=1.2•10^-7 m (Paschen series, infrared band)