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

Step 1: Recall the de Broglie Wavelength Formula

The de Broglie wavelength is given by:

λ = h / √(2mE)

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

Step 2: Express 1/λ² in Terms of E

Squaring both sides:

λ² = h² / 2mE

Taking the reciprocal:

1/λ² = 2mE / h²

Thus, 1/λ² is directly proportional to E.

Step 3: Conclude the Graph

A straight-line graph passing through the origin represents the relationship between 1/λ² and E.

Answer 2

Step 1: Recall the de Broglie Wavelength Formula

The de Broglie wavelength is given by:

λ = h / √(2mE)

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

Step 2: Express 1/λ² in Terms of E

Squaring both sides:

λ² = h² / 2mE

Taking the reciprocal:

1/λ² = 2mE / h²

Thus, 1/λ² is directly proportional to E.

Step 3: Conclude the Graph

A straight-line graph passing through the origin represents the relationship between 1/λ² and E.