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प्रश्न
If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is:
पर्याय
25
75
60
50
MCQ
उत्तर
75
Explanation:
As we know
λ = `"h"/"P" = "h"/sqrt(2"mK")` ....`(∵ "P" = sqrt(2"mK"))`
or `λ_1/λ_2 = sqrt("K"_2/"K"_1) = sqrt((16"K")/"K") = 4/1`
Therefore the percentage chanbe in de-Broglie wavelength
= `(1 - 4)/4 xx 100`
= − 75%
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