Question

The graph which shows the variation of \(\big(\frac{1}{\lambda^2} \big)\)and its kinetic energy, is \(E\) (where \(λ\) is de Broglie wavelength of a free particle):

• A.
• B.
• C.

  

• D.

  

Answer 1

The Correct Option is D

Using $\lambda = \frac{h}{\sqrt{2mE}}$, we find $\frac{1}{\lambda^2} \propto E$, yielding a linear graph.

Answer 2

Step 1: Recall the de Broglie Wavelength Formula 

The de Broglie wavelength is given by:

$$ \lambda = \frac{h}{\sqrt{2mE}} $$

• λ = de Broglie wavelength
• E = Kinetic energy
• m = Mass
• h = Planck’s constant

Step 2: Express \( \frac{1}{\lambda^2} \) in Terms of \( E \)

Squaring both sides:

$$ \lambda^2 = \frac{h^2}{2mE} $$

Taking the reciprocal:

$$ \frac{1}{\lambda^2} = \frac{2mE}{h^2} $$

Thus, \( \frac{1}{\lambda^2} \) is directly proportional to \( E \).

Step 3: Conclude the Graph

A straight-line graph passing through the origin represents the relationship between \( \frac{1}{\lambda^2} \) and \( E \).