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:(a) 25(b) 75(c) 60(d) 50
Answers
Answered by
1
Solution :
λ=h2mk−−−−√λ=h2mk
λ1=h2m×16k−−−−−−−√λ1=h2m×16k
λ=λ4λ=λ4
Percentage change in de- Broglie wave length,
λ−λ1λλ−λ1λ×100=(1−λ1λ)×100=(1−λ1λ)×100×100
=(1−14)=(1−14)×100=75%
λ=h2mk−−−−√λ=h2mk
λ1=h2m×16k−−−−−−−√λ1=h2m×16k
λ=λ4λ=λ4
Percentage change in de- Broglie wave length,
λ−λ1λλ−λ1λ×100=(1−λ1λ)×100=(1−λ1λ)×100×100
=(1−14)=(1−14)×100=75%
Similar questions