Question

If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is:
1. 25
2. 75
3. 60
4. 50

Answer 1

Correct Answer - Option 2 : 75

CONCEPT:

• De Broglie wavelength connects between the wavelength and momentum of the particle and is given by, and is given by 

\(⇒ λ = \frac{h}{P} = \frac{h}{mV}\)    ----(1)

Where m = mass of the particle and V = Velocity, h = Plancks constant 

Substituting the value \(P = \sqrt{2mE}\)in equation 1, it can be rewritten as

\(\Rightarrow λ = \frac{h}{\sqrt{2mE}}\)

Where m = mass, E= kinetic energy, 

CALCULATION :

Given E₂ = 2 E₁

• The kinetic energy of a particle in terms of wavelength can be written as

\(⇒ E=\dfrac{h^2}{2m\lambda^2}\)

The different kinetic energy can be written as 

\(⇒ E_{1}=\dfrac{h^2}{2m\lambda_{1}^2}\)

\(⇒ E_{2}=\dfrac{h^2}{2m\lambda_{2}^2}\)

The ratio  \(\frac{E_{1}}{E_{2}}\)can be written as

\(⇒ \dfrac{E_1}{E_2} = \dfrac{\lambda_2^2}{\lambda_1^2} \Rightarrow \dfrac{1}{16}= \dfrac{\lambda_2^2}{\lambda_1^2} \Rightarrow \dfrac{\lambda_2}{\lambda_1} = \dfrac{1}{4}\)

\(\Rightarrow\lambda_2 = \dfrac{\lambda_1}{4}\)

• The percentage change in wavelength can be expressed as

\(⇒ \Delta \lambda \% = \dfrac{\lambda_1 - \dfrac{\lambda_1}{4}}{\lambda_1}\times 100 = 75 \ \%\)

• Hence, option 2 is the answer