Question

٢٠ طاقة فوتون من ضوء أخضر أحادي اللون منبعث من مصدر ضوء عادي..........طاقة فوتون من ليزر الهيليوم - نيون.

• A. أقل من.
• B. أكبر من.
• C. نصف.
• D. تساوي.

Answer 1

Answer: (B)

أكبر من.

Answer 2

The energy of a photon ($E$) is directly proportional to its frequency ($\nu$) and inversely proportional to its wavelength ($\lambda$). The relationship is given by the formula:

$E = h\nu = \frac{hc}{\lambda}$

where:

• $h$ is Planck's constant
• $c$ is the speed of light in vacuum

To compare the energy of photons from different light sources, we need to compare their wavelengths.

1. Green Light: Monochromatic green light typically has a wavelength in the range of 495 nm to 570 nm. Let's consider an approximate average wavelength for green light, e.g., $\lambda_{green} \approx 520 \text{ nm}$.

2. Helium-Neon (He-Ne) Laser: A common He-Ne laser emits red light at a specific wavelength of $\lambda_{He-Ne} = 632.8 \text{ nm}$.

Comparison:

We compare the wavelengths:

$\lambda_{green} \approx 520 \text{ nm}$

$\lambda_{He-Ne} = 632.8 \text{ nm}$

Since $\lambda_{green} < \lambda_{He-Ne}$ (520 nm is less than 632.8 nm), and energy is inversely proportional to wavelength, the photon with the shorter wavelength will have higher energy.

Therefore, the energy of a photon from green light is greater than the energy of a photon from a Helium-Neon laser.