Question

Proton (P) and electron (e) will have same de-Broglie wavelength when the ratio of their momentum is (assume m_p = 1849me):

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

Hint:

For particles with equal de-Broglie wavelengths, their momenta must also be equal

Answer 1

The Correct Option is C

Step 1: Use the de-Broglie wavelength formula.- λ = $\frac{h}{p}$, where p is momentum.

- If λ_p = λ_e, then:

$\frac{h}{p_p} = \frac{h}{p_e}$ ⇒ $\frac{p_p}{p_e} = 1$.

Final Answer: The ratio of momentum is 1 : 1.

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