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

Guest Nov 9, 2020

#1**0 **

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

Guest Nov 9, 2020

#2**0 **

ugh so sorry (h times c divided by lambda)^2 does not equal (h divided by lambda times c)^2

Guest Nov 9, 2020

#3**0 **

*"(h times c divided by lambda)^2 does not equal (h divided by lambda times c)^2*"

This statement is complete nonsense!!!

Guest Nov 9, 2020

#4**0 **

(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

Guest Nov 9, 2020

#5**0 **

(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}

^{ }= h^{2}c^{2 }/ l^{2}

Guest Nov 9, 2020