PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
6th Edition
ISBN: 9781429206099
Author: Tipler
Publisher: MAC HIGHER
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Chapter 34, Problem 36P
(a)
To determine
The kinetic energy of a proton.
(b)
To determine
The kinetic energy of a proton.
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Chapter 34 Solutions
PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
Ch. 34 - Prob. 1PCh. 34 - Prob. 2PCh. 34 - Prob. 3PCh. 34 - Prob. 4PCh. 34 - Prob. 5PCh. 34 - Prob. 6PCh. 34 - Prob. 7PCh. 34 - Prob. 8PCh. 34 - Prob. 9PCh. 34 - Prob. 10P
Ch. 34 - Prob. 11PCh. 34 - Prob. 12PCh. 34 - Prob. 13PCh. 34 - Prob. 14PCh. 34 - Prob. 15PCh. 34 - Prob. 16PCh. 34 - Prob. 17PCh. 34 - Prob. 18PCh. 34 - Prob. 19PCh. 34 - Prob. 20PCh. 34 - Prob. 21PCh. 34 - Prob. 22PCh. 34 - Prob. 23PCh. 34 - Prob. 24PCh. 34 - Prob. 25PCh. 34 - Prob. 26PCh. 34 - Prob. 27PCh. 34 - Prob. 28PCh. 34 - Prob. 29PCh. 34 - Prob. 30PCh. 34 - Prob. 31PCh. 34 - Prob. 32PCh. 34 - Prob. 33PCh. 34 - Prob. 34PCh. 34 - Prob. 35PCh. 34 - Prob. 36PCh. 34 - Prob. 37PCh. 34 - Prob. 38PCh. 34 - Prob. 39PCh. 34 - Prob. 40PCh. 34 - Prob. 41PCh. 34 - Prob. 42PCh. 34 - Prob. 43PCh. 34 - Prob. 44PCh. 34 - Prob. 45PCh. 34 - Prob. 46PCh. 34 - Prob. 47PCh. 34 - Prob. 48PCh. 34 - Prob. 49PCh. 34 - Prob. 50PCh. 34 - Prob. 51PCh. 34 - Prob. 52PCh. 34 - Prob. 53PCh. 34 - Prob. 54PCh. 34 - Prob. 55PCh. 34 - Prob. 56PCh. 34 - Prob. 57PCh. 34 - Prob. 58PCh. 34 - Prob. 59PCh. 34 - Prob. 60PCh. 34 - Prob. 61PCh. 34 - Prob. 62PCh. 34 - Prob. 63PCh. 34 - Prob. 64PCh. 34 - Prob. 65PCh. 34 - Prob. 66PCh. 34 - Prob. 67PCh. 34 - Prob. 68PCh. 34 - Prob. 69PCh. 34 - Prob. 70PCh. 34 - Prob. 71PCh. 34 - Prob. 72PCh. 34 - Prob. 73P
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- A beam of 400 nm light is adjusted to have the same intensity as a beam of 600 nm light. How does the number of photons in the 400nm beam, N400 compare to the number of photons, N600 in the 600 nm beam. Question 5 options: N400 = 2*N600 N400 = 32600 N400 = N600 N400 = 23600 N400 = 2600 None of the other responses are correct.arrow_forwardWhat is the energy in eV and wavelength in µm of a photon that, when absorbed by a hydrogen atom, could cause a transition from the n = 4 to the n = 6 energy level? (a) energy in eV? (b) wavelength in µm?arrow_forwardCalculate the velocity (in m/s) of an electron that has a wavelength of 8.76 µm. m/s (b) Through what voltage (in V) must the electron be accelerated to have this velocity? (Assume the electron starts at rest.) Varrow_forward
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- The de Broglie wavelength of an electron has to do with spatial resolution of an electron microscope, which is often expressed in the unit of length Å (Angstrom). The 1 V potential difference causes an electron to gain kinetic energy EK of 1 electron Volt (eV). In the SI units, kinetic energy in eV must be converted to Joules. The conversion factor is 1 eV = 1.6 x 10-19 Joule. And, the formula for the wavelength is: λ = h / p = h / √(2 m EK) where m is electron mass. Calculate the de Broglie wavelength of an electron (in Å) when the electron is accelerated from rest through a potential difference of: a) 1 kV = 1,000 V (a low resolution setting of microscope), b) 10 kV = 10,000 V (intermediate resolution). c) 100 kV = 100,000 V (high resolution),arrow_forwardAssuming that 10.0% of a 100-W light bulb’s energy output is in the visible range (typical for incandescent bulbs) with an average wavelength of 580 nm, calculate the number of visible photons emitted per second.arrow_forwardBarium has a work function of 2.48 eV. What is themaximum kinetic energy of electrons if the metal is illuminated by UV light of wavelength 365 nm? What is their speed?arrow_forward
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