Question

For an electron and a proton (mp=1847 me) with same de-broglie wavelength, the ratio of linear momentum is equal to?

• A. 1:2
• B. 2:1847
• C. 1:1
• D. \(\sqrt{1847}{1}\)

Answer 1

The Correct Option is C

The correct option is (C): 1:1
\(\lambda=\frac{h}{p}\)
As \(p_{1} = p_{2}\)
\(\lambda_{1}=\lambda_{2}\)

Answer 2

The de Broglie wavelength is given by λ = h/p, where h is the Planck's constant and p is the momentum. Since the de Broglie wavelength is the same for both the electron and the proton, we can equate their respective momentum as:
\(p_{electron} = \frac{h}{λ}\)
\(p_{proton} = \frac{h}{λ}\)
Now, we can find the ratio of their linear momentum:
\(\frac{p_{proton}}{p_{electron}} = \frac{\frac{h}{λ}}{\frac{h}{λ}}\) = \(\frac{1}{1}\) = 1:1.
Therefore, the ratio of linear momentum between the electron and the proton is 1:1.
Answer. C