1) Calculate the energy, in units of Joules and eV, and momentum, in kg*m/s and eV/c,...

1) Calculate the energy, in units of Joules and eV, and momentum, in kg*m/s and eV/c, for a photon with wavelengths of: a) 1.5 pm, b) 120 nm, and c) 21.4 cm.

Expert Solution

genius_generous answered 7 months ago

Since force is m*a, it has units of kg m/s^2 . Energy has units of kg...

Since force is m*a, it has units of kg m/s^2 . Energy has units of kg m^2/s^2. Thus it appears that by lazy dimensional analysis that a force times a distance will have units of energy. This turns out to be the formula to calculate work. Note also that kg m^2/s^2 looks like mv^2. Here if we use dimensional analysis, we would *almost* find the formula for kinetic energy. We'd be off by a factor of two since KE=1/2 mv^2....

1) Using M/S energy band diagrams, determine if Nickel (5.2 eV workfunction) or Indium (4.1 eV...

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1. Calculate the permeability of electrons with energy of 0.1 eV and 0.5 eV respectively when...

1. Calculate the permeability of electrons with energy of 0.1 eV and 0.5 eV respectively when penetrating a positional energy barrier of 1.0 eV in height and 100 pm in length by tunnel phenomenon. k Attachment image=[2m(V-E)]^1/2 T={1+(e^kL-e^-kL)^2/16ε(1-ε)}^-1 (ε= E/V) 2. Explain the meaning of tunneling covered in the above problem with the wave and explain the difference between classical and quantum mechanics based on it.

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Calculate the number of joules that can be obtained from the fissioning of 1 kg of...

Calculate the number of joules that can be obtained from the fissioning of 1 kg of Uranium-235 (U-235), assuming 198 MeV average energy release per fission. a) How much energy can be obtained from the U-235 in 1 kg of natural uranium? b) Calculate how much energy, in joules, can be obtained from burning 1 kg of coal. c) What is the total energy contained in the US resources of coal and U-235?

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a) An electron with 10.0 eV kinetic energy hits a 10.1 eV potential energy barrier. Calculate the penetration depth. b) A 10.0 eV proton encountering a 10.1 eV potential energy barrier has a much smaller penetration depth than the value calculated in (a). Why? c) Give the classical penetration depth for a 10.0 eV particle hitting a 10.1 eV barrier.

Data Table A Mass of Cart (kg) Impulse (N.s) Velocity (m/s) Momentum (N.s) Change in Momentum...

Data Table A Mass of Cart (kg) Impulse (N.s) Velocity (m/s) Momentum (N.s) Change in Momentum % Diff. Before After Before After 0.2695 +0.351 - 0.673 +0.659 Data Table B Mass of Cart + mass bar (kg) Impulse (N.s) Velocity (m/s) Momentum (N.s) Change in Momentum % Diff. Before After Before After 0.4695 +0.346 -0.377 +0.372 % Difference= 2×(Change in momentum-Impulse)(Change in momentum+Impulse)×100= Questions What are possible reasons why the change in momentum is different from the measured impulse?...

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