Question

(۱)
A 50kev photon is compton scattered
by quasi free electron. If the scattered
photon comes off at 45 degree. What is
its wavelength?​

Answer 1

Answer:

so the answer will we 95

Explanation:

please give me brain list mark

Answer 2

Answer:

Wavelength = 0.0255 nm

Explanation:

DATA=

E=50KeV

(convert into Joules)

E=50x10^3 x 1.6x10^-19J

λ=?

λ'=?

SOLUTION:

FOR λ=

[ E = hf => E= hc/λ ]

So,

λ=hc/E

(Place the values)

λ=6.63x10^-34 x 3x10^8/50x10^3x1.6x10^-19

λ=0.0248x10^-9 m

OR λ= 0.0248 nm

FOR λ'=

λ'-λ=h/mc (1-cosθ)

(Place the values)

λ'-0.0248 = 6.63x10^-34/9.1x10^-31 x 3x10^8 (1-cos45)

λ'-0.0248 = 0.0007x10^-9

λ'= 0.0007x10^-9 + 0.0248x10^-9

λ'= 0.0255x10^-9

OR λ' = 0.0255 nm