Question

4. Water absorbs photons of wavelength 3x10-6m, with close to 100% efficiency. How many photons are needed to heat 100ml of water from 300C to 900C? How does the wavelength of this radiation compare to that of an electron in an electron beam with a kinetic energy of 5 keV?

Answer #1

a)

Q = mcdT

m = volume x density = 100 mL x 1 g/mL = 100 g

c = 4.179 J/(g⋅K)

dT = 900 ∘C - 300 ∘C = 600 oC = 600 K

Q = mcdT

= 100 g x 4.179 J/(g⋅K) x 600 K

= 250740 J

Q = 250740 J

Therefore, enerhy of photons required = 250740 J

E = nhc / λ n = no of photons

Given that λ = 3 x10^{-6} m

Then,

250740 J = n x (6.626 x
10^{-34} J.s) (3 x 10^{8} m/s) / (3
x10^{-6} m)

n = 37841 x 10^{20} photons

Therefore,

37841 x 10^{20} photons are
needed to heat 100ml of water from 300C to 900C.

b) Given that

E = 5 keV =
5000 eV = 5000 x 1.602 x 10^{-19} J

E = hc / λ

λ = hc/E

Then,

λ = (6.626 x 10^{-34} J.s) (3 x
10^{8} m/s) / 5000 x 1.602 x 10^{-19} J

= 2.48 x 10^{-10}
m

λ = 2.48 x 10^{-10} m

wavlength 2.48 x 10^{-10} m is
less than wavelength 3x10^{-6} m

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