A proton of 300 nm is absorbed by a gas and then reemits two proton .one remitted photon has wavelength of remmited photon is
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1
Answer:
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.
So, the corret option is A
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