Extra Credit 33

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Solve (i.e., write reasonable step-by-step tutorials for the solutions) of these exercises from the text)

https://phys.libretexts.org/Bookshel...ves_(Exercise)

#57, #91, #138

To find the peak blackbody radiation given wavelength, or temperature, use the Wiens Displacement Formula

\begin{equation}
\lambda *T = (2.989E-3 )m*K
\end{equation}

(a) given a wavelength of 400nm, calculate the peak temperature of a blackbody.

\begin{equation}
T = \frac{(2.989E-3) m*K}{400nm * \frac{10E-9m}{1nm}}
\end{equation}

T=7245K

(b) calculate wavelength

\begin{equation}
\lambda = /frac{(2.989E-3) m*K}{T}
\end{equation}

wavelength is 3.62E-6 m

For the energy changes between transitions of electrons in the hydrogen atom, you can use

\begin{equation}
\Delta E = [\frac{1}{n_F^2}-\frac{1}{n_i^2}]*13.6eV
\end{equation}

(a) plugging in all of the variables. nf=4 ni=3

\begin{equation}
\Delta E= [\frac{1}{4^2}-\frac{1}{3^2}]*13.6eV
\end{equation}

\begin{equation}
\Delta E= -.66eV
\end{equation}

(b) using n=2 and n=1

\begin{equation}
\Delta E= [\frac{1}{2^2}-\frac{1}{1^2}]*13.6eV
\end{equation}

\begin{equation}
\Delta E= 10.2eV
\end{equation}

(c) using n=3 and n=inf

\begin{equation}
\Delta E= [\frac{1}{2^2}-\frac{0}{1^2}]*13.6eV
\end{equation}

\begin{equation}
\Delta E= 1.51eV
\end{equation}

The kinetic energy of a photoelectron is determined by the formula

\begin{equation}
\frac{hc}{\lambda}-\varphi
\end{equation}

Where
h is the Planck Constant
c is the speed of light
lambda is the wavelength
varphi is the work function associated with any substance.

Calculate (a) radiation emitted by a 100kW radio station broadcasting at 800kHz

\begin{equation}
6.626E-34*800kHz *\frac{1000Hz}{1kHz}-2.48eV
\end{equation}

(a) is equal to 2.479eV, or barely and different from the work function.

(b)

\begin{equation}
\frac{6.626E-34* 2.998E8}{633*\frac{10E-9m}{1nm}}-2.48eV
\end{equation}

The outcome eV is -.519eV

(c)

\begin{equation}
\frac{6.626E-34* 2.998E8}{434*\frac{10E-9m}{1nm}}-2.48eV
\end{equation}

=.38eV

Answer (i.e., write the answer to these problems, but no need to show work. You can if you prefer)

https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/Book%3A_University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/06%3A_Photons_and_Matter_Waves/6.0E%3A_6.E%3A_Photons_and_Matter_Waves_(Exercise)

#7, #14, #42

Question: If everything radiates electromagnetic energy, why can we not see objects at room temperature in a dark room?
Answer: Everything radiates electromagnetic energy, but we can’t see everything in the dark because they don’t radiate in the visible spectrum, and instead radiate in the Infrared spectrum. Household objects dont glow in the visible unless they are burning, or very hot.

#14
Question: Which aspects of the photoelectric effect cannot be explained by classical physics?
Answer: The photoelectric effect has 3 things that cannot be explained by classical physics, like the (1) absence of lag time, the (2) independence of the kinetic energy of photoelectrons on the intensity of radiation, and the (3) presence of a cut-off frequency.
(1) Lag time is the classically predicted need for there to be a build-up of energy for irradiated or ionized to be able to kick off electrons for low energy systems, however, this is not observed experimentally.
(2) In classical theory, photoelectrons absorb energy continuously, and thus a high intensity of radiation should mean high kinetic energy. However experimentally we see that the incident radiation, whether it is low, high, or equal to the value of the stopping potential is always the same.
(3) Different materials have a different cutoff potential and rely on a different frequency. In classical mechanics, this does not make sense because the KE of Photoelectrons should not depend on frequency and there should, in fact, be no cut-off point whatsoever.

#42
Question: If a particle is accelerating, how does this affect its de Broglie wavelength?
Answer: As a de Broglie particle speeds up, the wavelength decreases with the formula
\begin{equation}
\lambda = \frac{h}{\sqrt{2mK_e}}
\end{equation}