Answer:
Explanation:
Given
Photon A has twice the Energy of Photon B
i.e. [tex]E_a=2E_b[/tex]
de Broglie wavelength is given by
[tex]\lambda =\frac{h}{P}[/tex]
[tex]\frac{1}{\lmabda }=\frac{P}{h}[/tex]
and Energy of Proton is given by
[tex]E=\frac{hc}{\lambda }[/tex]
[tex]E=hc\times \frac{P}{h}[/tex]
[tex]E=Pc[/tex]
where P=momentum
c=velocity of Light
[tex]P=\frac{E}{c} [/tex]
Momentum of A [tex]P_a=\frac{E_a}{c}=\frac{2E_b}{c}[/tex]
Momentum of B is [tex]P_b=\frac{E_b}{c}[/tex]
Thus [tex]P_a>P_b[/tex]
For wavelength
[tex]\lambda _a=\frac{h}{P_a}[/tex]
[tex]\lambda _b=\frac{h}{P_b}[/tex]
since [tex]P_a>P_b[/tex] therefore
[tex]\lambda _a<\lambda _b[/tex]
