Calculate the kinetic energy of the proton

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SUMMARY

The kinetic energy of a proton confined in a one-dimensional potential well with a width of 2.847 x 10-14 m and a mass of 1.67 x 10-27 kg can be calculated using the formula KE = (h2 * n2) / (8 * m * L2). For the quantum number n = 1, the correct calculation yields a kinetic energy of approximately 0.252 MeV. Initial calculations presented in the forum discussion were incorrect, but upon verification, the value was confirmed as 0.252 MeV.

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  • Understanding of quantum mechanics principles, specifically potential wells
  • Familiarity with the Planck constant (h = 6.626 x 10-34 J·s)
  • Knowledge of kinetic energy formulas in quantum physics
  • Basic proficiency in unit conversion, particularly between joules and MeV
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  • Explore unit conversion techniques between joules and MeV for particle physics
  • Investigate higher quantum states and their corresponding kinetic energies in potential wells
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Homework Statement


proton (mass = 1.67x10-27kg) is confined in a one dimensional potential well of infinite height and width of 2.847 x10-14m. Calculate the kinetic energy of the proton in the state with quantum number 1 in MeV


Homework Equations


KE = ((h^2)*(n^2))/(8*m*L^2) = ans / 1.6E-13 = ans in Mev


The Attempt at a Solution


((6.6E-34^2)*(1^2))/(8*1.67x10-27*2.847 x10-14^2)=1.7368E-14/1.6E-13=.10855
this answer is wrong, I am not sure why or what I am doing wrong. please help
 
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when I did your calculation I got 4 x 10^-14/1.6 x 10^-13 = 0.252
Check your calculation again
 
I just checked my calculation again, I came up with 0.252 also. Thank you!
 

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