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# I asked a question on here and I believe the person who answered it messed up.

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I asked a question on here and I believe the person who answered it messed up. If I am wrong please enlighten me. E^2 = (P x C)^2 + (M + C^2)^2 This is the formula given and I asked to solve for mass(assuming it's mass is unknown). Pic related is the answer that was given to me. They claim it equals zero, but I believe they overlooked the fact that the second parenthesis(left to right) is h divided by lambda multiplied by c^2 while the first parenthesis is  h multiplied by c^2 divided by lambda.  h divided by lambda multiplied by c^2 and h multiplied by c^2 divided by lambda are not equivalent to each other and hence the deduction of one to the other should not result in zero. Correct me if I am wrong, or if I'm right, admit that light has rest mass. If you want to use numbers you can use wavelength= 685 nanometers (but you have to convert that to meters)

E = Energy E = h times c^2 divided by lambda

P = Momentum P = h divided by lambda

C = Speed of light

M = Rest mass

h = Planck's constant

lambda = wavelength Nov 9, 2020

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Sorry when I was describing the parentheses I accidentally said c^2 when i should of put it in parenthesis and then put ^2 but my claim still stands( and now in correct format) that (h times c divided by lambda) does not equal (h divided by lambda times c)^2

Nov 9, 2020
#2
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ugh so sorry (h times c divided by lambda)^2 does not equal (h divided by lambda times c)^2

Nov 9, 2020
#3
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"(h times c divided by lambda)^2 does not equal (h divided by lambda times c)^2"

This statement is complete nonsense!!!

Nov 9, 2020
#4
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(h times c divided by lambda)^2 does not equal (h divided by lambda times c)^2 because with (h divided by lambda times c)^2 you multiply the denominator and numerator by c while with (h times c divided by lambda)^2 you only multiply the top. It is not nonsense, it's simple order of operations PEMDAS

Nov 9, 2020
#5
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(h times c divided by lambda)^2 does not equal (h divided by lambda times c)^2 because with (h divided by lambda times c)^2 you multiply the denominator and numerator by c while with (h times c divided by lambda)^2 you only multiply the top. It is not nonsense, it's simple order of operations PEMDAS

NO!    You DON'T multiply the denominator by c     Sorry...you are mistaken... (hc/l)2      is exactly the same as    (h/l   *c)2

= h2c/ l2

Guest Nov 9, 2020
#6
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Oh you're totally right. It doesn't make sense anymore. sorry for wasting your time.

Nov 9, 2020