the wavelength of a thermal neutron at 300K is lambda.What is the wavelenght of the neutron when at 1200K???(express in terms of lambda)
2007-03-02 19:27:00 · 4 answers · asked by naughtyme12345 1 in Science & Mathematics ➔ Chemistry
The de broglie wavelenght =Planck's constant/momentum.
momentum^2 ~ temperature. So when the temperature increases to 4 times the original value, momentum doubles. So the wavelength halved.
2007-03-02 19:40:41 · answer #1 · answered by PS 1 · 0⤊ 0⤋
My guess would be that the wavelength is given by E = h*c/L, where h = planck's const and c = vel of light. The energy is proportional to kT, where k = Boltzman's const. Therefore energy increases linearly with absolute temp, and L is inversely proportional to E. Your temp has increased by a factor of 4, so L will be lower by a factor of 4:
L(1200ºK) = L(400ºK) / 4
2007-03-02 19:46:02 · answer #2 · answered by gp4rts 7 · 0⤊ 0⤋
when you multiply by 4 the absolute temperature the energy is multiplied by 4.
As the energy is proportional to the square of velocity, when the energy is increased by 4 the velocity is increased by 2
and as the wavelength is inversely connected to velocity the wavelength is divided by two
so new wavelength lambda/2
2007-03-02 19:52:47 · answer #3 · answered by maussy 7 · 0⤊ 0⤋
Oh dear, the answer is upto my throat but I can't get it out!
2007-03-02 19:32:12 · answer #4 · answered by Sami V 7 · 0⤊ 0⤋