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Question
Let p and E denote the linear momentum and energy of a photon. If the wavelength is decreased,
Options
both p and E increase
p increases and E decreases
p decreases and E increases
both p and E decrease
Solution
both p and E increase
From the de-Broglie relation, wavelength,
`λ = h/p` ....(1)
⇒ `p = h/λ`
Here, h = Planck's constant
p = momentum of electron
It is clear from the above equation that `p ∝ 1/λ` .
Thus, if the wavelength `(λ)` is decreased, then momentum `(p)` will be increase .
Relation between momentum and energy :
`p = sqrt(2mE)`
Here, E = energy of electron
m = mass of electron
Substituting the value of p in equation (1), we get :
`λ = h/sqrt(2mE)`
⇒ `sqrt(E) = h/(λsqrt(2m)`
⇒ `E = h^2/(2mλ^2`
Thus, on decreasing λ , the energy will increase .
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