A) 25
B) 75
C) 60
D) 50
सही उत्तर: B
डी-ब्रॉग्ली तरंगदैर्ध्य के लिए ${\lambda_1}\frac{h}{p}=\frac{h}{\sqrt{2mK}},$ .(i)
${\lambda_2}=\frac{h}{\sqrt{2m16K}}=\frac{h}{4\sqrt{2mK}}=\frac{{\lambda_1}}{4}$ (ii)
${\lambda_2}=$25%,of,${\lambda_1}$ तरंगदैर्ध्य में 75% परिवर्तन है।
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