If K.E of free electron is doubled, its de broglie wavelength becomes?

Explanation

The de Broglie wavelength (λ) of a particle is given by the equation λ = h / p, where h is Planck's constant and p is the momentum of the particle. The kinetic energy (K.E.) of an electron can be related to its momentum using the equation K.E. = p² / (2m), where m is the mass of the electron. Let's consider the initial kinetic energy of the electron as K.E.1. If the kinetic energy is doubled (2K.E.1), the new kinetic energy becomes K.E. 2 = 2K.E.1. Since the momentum (p) of the electron is related to its kinetic energy, we can write: K.E.1 = p₁² / (2m) K.E.2 = p₂² / (2m) Dividing the second equation by the first equation: 2K.E.1 / K.E.1 = (p₂² / (2m)) / (p₁² / (2m)) 2 = (p₂² / p₁²) Taking the square root of both sides: √2 = p₂ / p₁ Now, we can substitute this relationship between the momenta into the de Broglie wavelength equation: λ₂ = h / p₂ = h / (p₁ * √2) Rearranging, we get: λ₂ = (1/√2) * (h / p₁) = (1/√2) * λ₁ Therefore, when the kinetic energy of a free electron is doubled, its de Broglie wavelength becomes (1/√2) times the original wavelength.