Question

The momenta of a proton, a neutron and an electron are in the ratio 3:2:1, then their respective de Broglie wavelengths are in the ratio

• A. 1:1:1
• B. 2:3:6
• C. 1:2:3
• D. 6:3:2
• E. 4:2:1

Answer 1

The Correct Option is B

The de Broglie wavelength \( \lambda \) is related to the momentum \( p \) of a particle by the equation: \[ \lambda = \frac{h}{p} \] where \( h \) is Planck's constant and \( p \) is the momentum of the particle. Given that the momenta of the proton, neutron, and electron are in the ratio 3:2:1, their respective de Broglie wavelengths will be inversely proportional to their momenta. This means the particle with the highest momentum will have the shortest wavelength. Thus, the de Broglie wavelengths will be in the ratio: \[ \lambda_{\text{proton}} : \lambda_{\text{neutron}} : \lambda_{\text{electron}} = \frac{1}{3} : \frac{1}{2} : 1 \] Simplifying this ratio, we get: \[ \lambda_{\text{proton}} : \lambda_{\text{neutron}} : \lambda_{\text{electron}} = 2 : 3 : 6 \]

The correct option is (B) : \(2:3:6\)

Answer 2

The de Broglie wavelength is given by:

\(\lambda = \frac{h}{p}\)

where:
    \(\lambda\) is the de Broglie wavelength
    h is Planck’s constant (constant for all particles)
    p is the momentum of the particle

Hence, \(\lambda \propto \frac{1}{p}\)

Given: p_proton : p_neutron : p_electron = 3 : 2 : 1
So the wavelengths will be inversely proportional:

\(\lambda_{proton} : \lambda_{neutron} : \lambda_{electron} = \frac{1}{3} : \frac{1}{2} : 1\)

Multiply all by 6 to remove denominators:

\(2 : 3 : 6\)

Correct Option: 2 : 3 : 6