if the kinetic energy of a free electron double its deBroglie wavelength change by the factor is
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Ans : (c)
λ=hmvandK=12mv2=(mv)22mλ=hmvandK=12mv2=(mv)22m
⇒mv=2mK−−−−√⇒mv=2mK
So, λ=h2mK−−−−−√λ=h2mK
⇒λ∼1K−−√⇒λ∼1K
λ2λ1=K1−−−√K2−−−√=K1−−−√2K1−−−−√(K2=2K1)λ2λ1=K1K2=K12K1(K2=2K1)
⇒λ2λ1=12–√
λ=hmvandK=12mv2=(mv)22mλ=hmvandK=12mv2=(mv)22m
⇒mv=2mK−−−−√⇒mv=2mK
So, λ=h2mK−−−−−√λ=h2mK
⇒λ∼1K−−√⇒λ∼1K
λ2λ1=K1−−−√K2−−−√=K1−−−√2K1−−−−√(K2=2K1)λ2λ1=K1K2=K12K1(K2=2K1)
⇒λ2λ1=12–√
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Hey there! Your answer is 1/root2 which is the change.
You can use the formula lamda=h/root over of 2mk
You can use the formula lamda=h/root over of 2mk
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