Question

A particle of mass $9.1 \times 10^{-31}$ kg travels in a medium with a speed of $10^6$ m/s and a photon of a radiation of linear momentum $10^{-27}$ kg m/s travels in vacuum. The wavelength of photon is _________ times the wavelength of the particle.

Hint:

The de Broglie wavelength formula $\lambda = h/p$ applies to both matter particles and photons. This allows for a direct comparison of their wavelengths if their momenta are known.

Answer 1

Correct Answer: 910

First, we calculate the de Broglie wavelength of the particle ($\lambda_{particle}$).
The formula for de Broglie wavelength is $\lambda = \frac{h}{p}$, where p is the momentum ($p=mv$).
Given: mass of particle, $m = 9.1 \times 10^{-31}$ kg, and speed, $v = 10^6$ m/s.
Momentum of the particle, $p_{particle} = mv = (9.1 \times 10^{-31} \text{ kg}) \times (10^6 \text{ m/s}) = 9.1 \times 10^{-25}$ kg m/s.
$\lambda_{particle} = \frac{h}{p_{particle}} = \frac{h}{9.1 \times 10^{-25}}$.
Next, we find the wavelength of the photon ($\lambda_{photon}$).
The formula for the wavelength of a photon is also $\lambda = \frac{h}{p}$.
Given: momentum of photon, $p_{photon} = 10^{-27}$ kg m/s.
$\lambda_{photon} = \frac{h}{p_{photon}} = \frac{h}{10^{-27}}$.
Finally, we find the ratio of the photon's wavelength to the particle's wavelength.
$\frac{\lambda_{photon}}{\lambda_{particle}} = \frac{h/p_{photon}}{h/p_{particle}} = \frac{p_{particle}}{p_{photon}}$
$\frac{\lambda_{photon}}{\lambda_{particle}} = \frac{9.1 \times 10^{-25} \text{ kg m/s}}{10^{-27} \text{ kg m/s}}$
$\frac{\lambda_{photon}}{\lambda_{particle}} = 9.1 \times 10^{(-25 - (-27))} = 9.1 \times 10^2 = 910$.
So, the wavelength of the photon is 910 times the wavelength of the particle.