COMEDK Exam 2017 Solved Paper

Given, initial value of de-Broglie wavelength,
λ1=1‌nm=1×10−9m
Final value of de-Broglie wavelength,
λ2=0.5‌nm=0.5×10−9m
de-Broglie wavelength in terms of energy E is given as
‌λ=‌

h

√2mE

∴‌‌λ∝‌

1

√E

⇒‌‌‌

λ1

λ2

‌=√‌

E2

E1

where, E1 and E2 are initial and final energy,
‌⇒‌‌‌

1×10−9

0.5×10−9

=√‌

E2

E1

‌⇒‌‌2=√‌

E2

E1

‌⇒‌‌E2=4E1
‌∴‌ Addition of energy ‌=E2−E1=4E1−E1=3E1
‌‌ = thrice the initial energy ‌
‌