JEE Main 26-Aug-2021 Shift 2 Solved Paper

Given, initial kinetic energy, E1=E
Initial de-Broglie wavelength, λ1=λ
Consider the wavelength of particle changes to 75% of λ after
providing energy ∆E to the particle.
Hence, final wavelength of particle, λ2=0.75λ
Final energy, E2=E+ΔE
Relationship between de-Broglie wavelength and energy of particleis given as
λ=

h

√2mE

Here, h is Planck’s constant and m is mass of particle which is alsoconstant term.
Therefore, we get
λ∝

1

√E

Thus, we can write the relationship as

λ1

λ2

=

√E2

√E1

⇒

λ

0.75λ

=

√E+ΔE

√E

⇒

4

3

=

√E+ΔE

√E

Squaring both sides of above equation, we get
⇒

16

9

=

E+ΔE

E

⇒16E=9E+9ΔE
ΔE=

7

9

E