Question

An electron and a photon each have the same de-Broglie wavelength of 1.0 nm. The ratio of their linear momenta is

• A. 1:1
• B. 1:2
• C. 2:1
• D. 4:9

Hint:

Both use p = h/wavelength, so equal wavelength gives equal momentum.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
Both particles have the same wavelength, 1.0 nm. We must compare their momenta as a ratio.

Step 2: Key Formula or Approach:
The de-Broglie relation for any particle, and also the relation for a photon, is \[ \lambda = \frac{h}{p} \quad \Rightarrow \quad p = \frac{h}{\lambda} \]

Step 3: Momentum of the Electron:
\[ p_e = \frac{h}{\lambda} \]

Step 4: Momentum of the Photon:
For a photon, \(E = hc/\lambda\) and \(p = E/c = h/\lambda\). So \[ p_{ph} = \frac{h}{\lambda} \]

Step 5: Take the Ratio:
\[ \frac{p_e}{p_{ph}} = \frac{h/\lambda}{h/\lambda} = 1 \] The ratio is 1:1. The mass and speed of the two particles are very different, but momentum depends only on wavelength.

Step 6: Checking Each Option:
Option 1 (1:1) is correct. Options 2, 3 and 4 would need the momenta to differ, which the relation \(p = h/\lambda\) does not allow when \(\lambda\) is equal.

Final Answer:
Same wavelength means same momentum, so the ratio is 1:1. This is option 1. \[ \boxed{1:1} \]