The energy that should be added to an electron to reduce its de-Broglie wavelength from 1 nm to 0.5 nm is

Question

The energy that should be added to an electron to reduce its de-Broglie wavelength from 1 nm to 0.5 nm is (A) four-times the initial energy (B) equal to the initial energy (C) two-times the initial energy (D) three-times the initial energy

Tardigrade Admin · Accepted Answer

We know that, We have, λ=(h/√2 m k) ∴ Kinetic energy (K E) propto (1/λ2) Hence, (K E2/K E1)=((λ1/λ2))2 Here, λ1 =1 nm λ2 =0.5 nm (K E2/K E1) =((1/0.5))2 K E2 =4 K E1 Hence, the kinetic energy is increases three times. Δ K E=3 K E1

Q. The energy that should be added to an electron to reduce its de-Broglie wavelength from 1nm to 0.5nm is

four-times the initial energy

equal to the initial energy

two-times the initial energy

three-times the initial energy

We know that, We have, λ=2mk​h​ ∴ Kinetic energy (KE)∝λ21​ Hence, KE1​KE2​​=(λ2​λ1​​)2 Here, λ1​=1nm λ2​=0.5nm KE1​KE2​​=(0.51​)2 KE2​=4KE1​ Hence, the kinetic energy is increases three times. ΔKE=3KE1​

Q. The energy that should be added to an electron to reduce its de-Broglie wavelength from $1 nm$ to $0.5 nm$ is