ILLUSTRATION 4.41
A photon of 300 nm is absorbed by a gas, which then re-emits
two photons. One re-emitted photon has a wavelength of 400
nm. Calculate the energy of the other photon re-emitted out.
Answers
Answer:
Explanation:
1/LAMDA=1/LAMDA1+1/LAMDA2
1/300=1/400+1/LAMDA2
1/LAMDA2=1/300-1/400
1/LAMDA2=100/120000
LAMDA2=1200nm
energy=hc/lamda
=6.67*10^-28*3*10^8/1200*10-9
=0.016*10^-11
Explanation:
This symmetric loving nature of Nature gave rise to de Broglie relation.
The de Broglie equation relates a moving particle's wavelength with its momentum. The de Broglie wavelength is the wavelength, λ, associated with a massive particle and is related to its momentum, p, through the Planck constant, h
λ =
p
h
In other words, you can say that matter also behaves like waves. This is what is called as de Broglie hypothesis.
E=
λ
hc
wavelength of 1
st
photon = λ
1
wavelength of 2
nd
photon = λ
2
E(total) = E1 + E2 = Emitted energy
λ
hc
=
λ
1
hc
+
λ
2
hc
300
1
=
496
1
+
λ
2
1
λ
2
= 759nm = wavelength of second photon.
Hope it would be helpful to you
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