Q#1
The negative muon has a charge equal to that of an electron but a mass that is 207 times as great. Consider a hydrogenlike atom consisting of a proton and a muon.
(a) What is the reduced mass of the atom?
(b) What is the ground-level energy (in electron volts)?
(c) What is the wavelength of the radiation emitted in the transition from the level to the level?
Answer:
(a) We find the reduced atomic mass by using
$m_r=\frac{m_1m_2}{m_1+m_2}$
$m_r=\frac{207m_em_p}{207m_e+m_p}$
$m_r=\frac{207(9.109 \times 10^{-31} \ kg)(1.673 \times 10^{-27} \ kg)}{207(9.109 \times 10^{-31} \ kg) + (1.673 \times 10^{-27} \ kg)}$
$m_r=1.69 \times 10^{-28} \ kg$
We have used $m_e$ to denote the electron mass.
(b) We use
$E_n=-\frac{1}{\epsilon_0^2}\frac{m_re^4}{8n^2h^2}$
Write as
$E_n=\left(\frac{m_r}{m_H}\right)\left(-\frac{1}{\epsilon_0^2}\frac{m_He^4}{8n^2h^2}\right)$
since we know that
$\frac{1}{\epsilon_0^2}\frac{m_He^4}{8h^2} =13,6 \ eV$
Here $m_H$ denotes
the reduced mass for the hydrogen atom;
$m_H=0,99946(9.109 \times 10^{-31} \ kg)=9,104 \times 10^{-31} \ kg$, so
$E_n=\left(\frac{m_r}{m_H}\right)\left(-\frac{13,6 \ eV}{n^2}\right)$
$E_1=\left(\frac{1.68 \times 10^{-28} \ kg}{9.109 \times 10^{-31} \ kg}\right)\left(-\frac{13,6 \ eV}{n^2}\right)$
$E_1=(186)(-13,6 \ eV)=-2,53 \ keV$
(c) From part (b), $E_n=\left(\frac{m_r}{m_H}\right)\left(-\frac{R_H ch}{n^2}\right)$
where $R_H=1.097 \times 10^7 \ m^{-1}$ is the Rydberg constant for the hydrogen atom. Use this result in $\frac{hc}{\lambda}=E_i-E_f$ to find an expression for $\frac{1}{\lambda}$
The initial level for the transition is the $n_i=2$ level and the final level is the $n_f=1$ level
$\frac{hc}{\lambda}=\frac{m_r}{m_H}\left(-\frac{R_H ch}{n_i^2}-\left(-\frac{R_H ch}{n_f^2}\right)\right)$
$\frac{1}{\lambda}=\frac{m_r}{m_H}R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$
$\frac{1}{\lambda}=\frac{1.67 \times 10^{-28} \ kg}{9.109 \times 10^{-31} \ kg}(1.097 \times 10^7 \ m^{-1})\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
$\frac{1}{\lambda}=1.527 \times 10^9 \ m^{-1}$
$\lambda = 0.655 \ nm$
Q#2
An atom with mass m emits a photon of wavelength
(a) What is the recoil speed of the atom?
(b) What is the kinetic energy K of the recoiling atom?
(c) Find the ratio K/E where E is the energy of the emitted photon.
If this ratio is much less than unity, the recoil of the atom can be neglected in the emission process. Is the recoil of the atom more important for small or large atomic masses? For long or short wavelengths?
(d) Calculate K (in electron volts) and K/E for a hydrogen atom (mass $1,67 \times 10^{-27} \ kg$) that emits an ultraviolet photon of energy 10.2 eV. Is recoil an important consideration in this emission process?
Answer:
Apply conservation of momentum to the system of atom and emitted photon.
Assume the atom is initially at rest. For a photon $E=\frac{hc}{\lambda}$ and $p = \frac{h}{\lambda}$
(a) Assume a non-relativistic velocity and conserve momentum
$mv=\frac{h}{\lambda}⇒v=\frac{h}{m \lambda}$
(b) the kinetic energy K of the recoiling atom is given by
$K = \frac{1}{2}mv^2$
$K = \frac{1}{2}m\left(\frac{h}{m \lambda}\right)^2$
$K = \frac{h^2}{2m \lambda^2}$
(c) the ratio K/E where E is the energy of the emitted photon is given by
$\frac{K}{E}=\frac{\frac{h^2}{2m \lambda^2}}{\frac{hc}{\lambda}}$
$\frac{K}{E}=\frac{h^2}{2m \lambda^2} \times \frac{\lambda}{hc}$
$\frac{K}{E}=\frac{h}{2mc \lambda^2}$
Recoil becomes an important concern for small m and small λ since this ratio becomes large in those limits.
(d) $E = 10 \ eV$
$\lambda = \frac{hc}{E}$
$\lambda = \frac{(6.63 \times 10^{-34} \ J.s)(3.00 \times 10^8 \ m/s)}{10 \ eV}$
$\lambda = 1.22 \times 10^{-7} \ m = 122 \ nm$
$K = \frac{(6.63 \times 10^{-34} \ J.s)^2}{2(1.67 \times 10^{-27} \ kg)(1.22 \times 10^{-7} \ m)^2}$
$K=8.84 \times 10^{-27} \ J = 5.53 \times 10^{-8} \ eV$
$\frac{K}{E}=\frac{5.53 \times 10^{-8} \ eV }{10.2 \ eV}=5.42 \times 10^{-9}$. This is quite small so recoil can be neglected
For emission of photons with ultraviolet or longer wavelengths the recoil kinetic energy of the atom is much less than the energy of the emitted photon.
Q#3
(a) What is the smallest amount of energy in electron volts that must be given to a hydrogen atom initially in its ground level so that it can emit the $H_{alpha}$line in the Balmer series?
(b) How many different possibilities of spectral-line emissions are there for this atom when the electron starts in the level and eventu[1]ally ends up in the ground level? Calculate the wavelength of the emitted photon in each case.
Answer:
The $H_{\alpha}$ line in the Balmer series corresponds to the 3 n = to 2 n = transition.
$E_n=-\frac{13.6 \ eV}{n^2}.\frac{hc}{\lambda}=\Delta E$
The Hα line in the Balmer series corresponds to the 3 n = to 2 n = transition.
$E_3-E_1=(-13.6 \ eV)\left(\frac{1}{3^2}-\frac{1}{1^2}\right)=12.1 \ eV$
(b) There are three possible transitions n = 3 → n = 1: ΔE = 12.1 eV and
$\lambda = \frac{(6.63 \times 10^{-34} \ J.s)(3.00 \times 10^8 \ m/s)}{(12.1 \ eV)(1.60 \times 10^{-19} \ C)}=103 \ nm $
n = 3 → n = 2: $E_3-E_1=(-13.6 \ eV)\left(\frac{1}{3^2}-\frac{1}{2^2}\right)=1.89 \ eV$ and
$\lambda = \frac{(6.63 \times 10^{-34} \ J.s)(3.00 \times 10^8 \ m/s)}{(1.89 \ eV)(1.60 \times 10^{-19} \ C)}=657 \ nm $
n = 2 → n = 1: $E_2-E_1=(-13.6 \ eV)\left(\frac{1}{2^2}-\frac{1}{1^2}\right)=10.2 \ eV$ and
$\lambda = \frac{(6.63 \times 10^{-34} \ J.s)(3.00 \times 10^8 \ m/s)}{(10.2 \ eV)(1.60 \times 10^{-19} \ C)}=122\ nm $
The larger the transition energy for the atom, the shorter the wavelength.
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