Find the de Broglie wavelength A for an electron moving at a speed of 1.00 × 106 m/s. (Note that this speed is low enough that the classical momentum formula p = mv is still valid.) Recall that the mass of an electron is me = 9.11 × 10-³1 kg, and Planck's constant is h = 6.626 × 10-34 J.s.
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- : In class we did a problem where we saw that Planck’s law, which is given byI(λ, T) = 2πhc2λ5(ehc/λkbT − 1),at high wavelengths reduces to the classical predictionI(λ, T) = 2πckBTλ4.We did this by using the MacLaurin series for an exponential.1 + x +x22! +x33! +x44! + ...When we did this problem in class, we reasoned that,when the wavelength (λ) is large, the term hc/λkbTis small enough that any term (hc/λkbT)2 or (hc/λkbT)3 or any higher power is neglible. That’s how weshowed that, in the limit of large λ, Planck’s law reduces to the classical prediction. Now assume that wewant to make a slightly better approximation. We still assume that λ is large and therefore (hc/λkbT)3 and(hc/λkbT)4 and all higher powers are negligible, but now we want to work at the level of precision where(hc/λkbT)2is not negligible. What does Planck’s law reduce to in this case?In quantum interpretation of the electromagnetic waves in vacuum the photon has the energy E = hw/2n and the momentump= hk/2n. So the ratio E of the energy to the momentum is = c, the speed of light in k vacuum. Similar relation can be obtained in classical electrodynamics as follows. Consider the time-averaged energy density of the electromagnetic field ɛ = B² /2µ0 + €0 E² /2 for a plane wave propagating in vacuum along the z-direction. Calculate the time-averaged energy flux (Poynting flux) for such a wave S = E × H and confirm that its ratio to the time-averaged energy density ratio is equal to the speed of light in vacuum, S/e = c.(a) A vacuum photocell is sequentially illuminated with light of different wavelengths 2. A voltmeter is used to determine that there is a different voltage between the cathode and the anode. V (iii) Determine a relation for Planck's constant in terms of pairs of voltage measurements at different wavelengths such that W₁ cancels out. (iv) Evaluate Planck's constant for the following pair of measurements: measurement 1 finds = 447 nm and V=635 mV, and measurement 2 finds = : 502 nm and V=339 mV.
- Planck’s constant has the value h = 6.626 × 10–34 joule-seconds (J-s), and the speed of light is c = 3 × 108 m/s. Using these values, calculate the wavelength carried by photons emitted with an energy of 1.1 × 10-19 J.What is the energy of a photon if it's frequency is 1.00s-1?. Find the momentum of a photon in eV/c and in kg·m/s if the wavelength is (a) 400 nm, (b) 1 Å = 0.1 nm, (c) 3 cm, and (d) 2 nm.
- 3.1. What is the de Broglie wavelength of an electron that has been accelerated through a potential difference of AV = 150 V?A) Find the wavelength of a photon that has energy of 27 eVeV. Express your answer with the appropriate units. B) Find the wavelength of an electron that has energy of 27 eVeV. (The energy of the electron is its kinetic energy.) Express your answer with the appropriate units.i know the answer is NOT 9.42(10^-21)
- What is the wavelength in meters of a photon generated by an electron going from n = (8.000x10^0) to n= (1.00x10^0)? Answer to 3 significant figures and in scientific notation. Sorry about the two n values, I can either do scientific notation need for the answer or you have to enter the answer without scientific notation. I can't change between them. Note: Your answer is assumed to be reduced to the highest power possible. Your Answer: Answer x10 unitsWhat is the wavelength of a photon emitted when an electron jumps from the n=3 to the n=2 energy levels of a lithium atom (Z=3)? Express your answer in nanometers and keep three significant digits.