00:01
But a of the given problem, from a d -brogly -wevelinth, we can write lambda is equal to h divided by p.
00:06
P is the momentum and h is a point constant.
00:10
We can replace p with a relativistic definition of momentum that will give us h times 1 minus v squared divided by c -square, 1 over 2 divided by m -v.
00:22
From here then we can write lambda square m square v square is equal to h taking square on both sides h square one minus v square divide by c square which will be equal to h square uh edge square v square so minus sign here c square we multiplied h inside um then solving further lambda square m square v square plus h v squared divide by c square is equal to h square.
00:58
From here we solve for a v.
01:03
So we make a v here on left -hand side, which is v square is equal to h squared divided by lambda square, m square, plus h square, divide by c square, which we can write in c squared divided by lambda square, m square, c squared, divided by h square, plus one.
01:23
Taking square root, we get v to v .h divided by 1xm, mc lambda divided by h, square whole 1 over 2...