Question

Photon A has twice the energy of photon B.

Is the momentum of A less than, equal to, or greater than that of B?
Less Equal Greater Is the wavelength of A less than, equal to, or greater than that of B?
Less
Greater
Equal

Answer 1

Step-by-step explanation:

Given

Photon A has twice the Energy of Photon B

i.e.
`E_a=2E_b`

de Broglie wavelength is given by

`\lambda =(h)/(P)`

`(1)/(\lmabda )=(P)/(h)`

and Energy of Proton is given by

`E=(hc)/(\lambda )`

`E=hc* (P)/(h)`

`E=Pc`

where P=momentum

c=velocity of Light

`P=(E)/(c)`

Momentum of A
`P_a=(E_a)/(c)=(2E_b)/(c)`

Momentum of B is
`P_b=(E_b)/(c)`

Thus
`P_a>P_b`

For wavelength

`\lambda _a=(h)/(P_a)`

`\lambda _b=(h)/(P_b)`

since
`P_a>P_b` therefore

`\lambda _a<\lambda _b`